GEOMETRICAL MEANING OF A DERIVATIVE
The essence of calculus is the derivative. The derivative is the instantaneous rate of change of a function with respect to one of its variables. This is equivalent to finding the slope of the tangent line to the function at a point.
Let \(y=f(x)\) be a function of \(x\). Let \(\Delta y\) be the change in \(y\) corresponding to a small change \(\Delta x\) in \(x\). Then, \(\frac{\Delta y}{\Delta x}\) represents the change in \(y\) due to a unit change in \(x\).
In other words, \(\frac{\Delta y}{\Delta x}\) represents the average rate of change of \(y\) w.r.t. \(x\) as \(x\) changes from \(x\) to \(x+\Delta x\).
As \(\Delta x \rightarrow 0\), the limiting value of this average rate of change of \(y\) with respect to \(x\) in the interval \([x, x+\Delta x]\) becomes the instantaneous rate of change of \(y\) w.r.t. \(x\).
Thus,
\(
\begin{aligned}
& \lim _{\Delta x \rightarrow 0} \frac{\Delta y}{\Delta x}=\text { Instantaneous rate of change of } y \text { w.r.t. } x \\
& \Rightarrow \frac{d y}{d x}=\text { Rate of change of } y \text { w.r.t. } x\left[\because \lim _{\Delta x \rightarrow 0} \frac{\Delta y}{\Delta x}=\frac{d y}{d x}\right]
\end{aligned}
\)
The word “instantaneous” is often dropped.
Hence, \(\frac{d y}{d x}\) represents the rate of change of \(y\) w.r.t. \(x\) for a definite value of \(x\).
wherever the limit exists is defined to be the derivative of \(f(x)\) at \(x\) and is denoted by \(f^{\prime}(x)[\frac{d f(x)}{d x}]\). This definition of derivative is also called the first principle of derivative. Thus \(\quad f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\)
Illustration: Find the derivative \(e^x\)
Solution: \(f^{\prime}(x)=\lim _{h \rightarrow 0}[f(x+h)-f(x)] / h\)
\(e^{x+h}=e^x \cdot e^h\)
\(\lim _{x \rightarrow 0}\left(e^x-1\right) / x=1\)
Using the above formulas, we have
\(
\begin{aligned}
& d\left(e^x\right) / d x=\lim _{h \rightarrow 0}\left[e^{x+h}-e^x\right] / h \\
& =\lim _{h \rightarrow 0}\left[e^x \cdot e^h-e^x\right] / h \\
& =\lim _{h \rightarrow 0} e^x\left[e^h-1\right] / h \\
& =e^x \lim _{h \rightarrow 0}\left[e^h-1\right] / h \\
& =e^x \times 1 \\
& =e^x
\end{aligned}
\)
Example 1: Find the derivative of \(e^{\sqrt{x}}\) w.r.t. \(x\) using the first principle.
Solution:
\(
\begin{aligned}
& \text { Let } f(x)=e^{\sqrt{x}}, \text { then } f(x+h)=e^{\sqrt{x+h}} \\
& \frac{d}{d x}(f(x))=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}
\end{aligned}
\)
\(
\begin{aligned}
& =\lim _{h \rightarrow 0} \frac{e^{\sqrt{x+h}}-e^{\sqrt{x}}}{h} \\
& =e^{\sqrt{x}} \lim _{h \rightarrow 0}\left(\frac{e^{\sqrt{x+h}-\sqrt{x}}-1}{h}\right) \\
& =e^{\sqrt{x}} \lim _{h \rightarrow 0}\left(\frac{e^{\sqrt{x+h}-\sqrt{x}}-1}{\sqrt{x+h}-\sqrt{x}}\right)\left(\frac{\sqrt{x+h}-\sqrt{x}}{h}\right) \\
& =e^{\sqrt{x}} \lim _{h \rightarrow 0}\left(\frac{e^{\sqrt{x+h}-\sqrt{x}}-1}{\sqrt{x+h}-\sqrt{x}}\right) \times \\
& \frac{(\sqrt{x+h}-\sqrt{x})(\sqrt{x+h}+\sqrt{x})}{h(\sqrt{x+h}+\sqrt{x})}
\end{aligned}
\)
\(
\begin{aligned}
= & e^{\sqrt{x}} \lim _{h \rightarrow 0}\left(\frac{e^y-1}{y}\right) \lim _{h \rightarrow 0} \frac{x+h-x}{h(\sqrt{x+h}+\sqrt{x})} \\
& \text { where } y=\sqrt{x+h}-\sqrt{x}(\because \text { when } h \rightarrow 0, y \rightarrow 0) \\
= & e^{\sqrt{x}} \times 1 \times\left(\frac{1}{\sqrt{x}+\sqrt{x}}\right)=\frac{e^{\sqrt{x}}}{2 \sqrt{x}}
\end{aligned}
\)
Example 2: If \(f(x)=x \tan ^{-1} x\), find \(f^{\prime}(\sqrt{3})\) using the first principle.
Solution: We have \(f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\)
\(
\begin{aligned}
& \therefore f^{\prime}(\sqrt{3})=\lim _{h \rightarrow 0} \frac{f(\sqrt{3}+h)-f(\sqrt{3})}{h} \\
& =\lim _{h \rightarrow 0} \frac{(\sqrt{3}+h) \tan ^{-1}(\sqrt{3}+h)-\sqrt{3} \tan ^{-1} \sqrt{3}}{h} \\
& =\lim _{h \rightarrow 0} \frac{\sqrt{3}\left[\tan ^{-1}(\sqrt{3}+h)-\tan ^{-1} \sqrt{3}\right]+h \tan ^{-1}(\sqrt{3}+h)}{h} \\
& =\lim _{h \rightarrow 0} \frac{\sqrt{3}}{h} \tan ^{-1}\left(\frac{\sqrt{3}+h-\sqrt{3}}{1+\sqrt{3}(\sqrt{3}+h)}\right)+\lim _{h \rightarrow 0} \tan ^{-1}(\sqrt{3}+h) \\
& =\sqrt{3} \lim _{h \rightarrow 0}\left\{\frac{\tan ^{-1}\left(\frac{h}{4+\sqrt{3} h}\right)}{\frac{h}{4+\sqrt{3} h}}\right\} \frac{1}{4+\sqrt{3} h}+\lim _{h \rightarrow 0} \tan ^{-1}(\sqrt{3}+h) \\
& \Rightarrow f^{\prime}(\sqrt{3})=\sqrt{3} \times 1 \times \frac{1}{4}+\tan ^{-1} \sqrt{3} \\
& =\frac{\sqrt{3}}{4}+\tan ^{-1} \sqrt{3}
\end{aligned}
\)
Example 3: Find the derivative of \(\sqrt{4-x}\) w.r.t. \(x\) using the first principle.
Solution: Let \(f(x)=\sqrt{4-x}\), then \(f(x+h)=\sqrt{4-(x+h)}\)
\(
\begin{aligned}
& \therefore \frac{d}{d x}(f(x))=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\
&=\lim _{h \rightarrow 0} \frac{\sqrt{4-(x+h)}-\sqrt{4-x}}{h} \\
&=\lim _{h \rightarrow 0} \frac{\{\sqrt{4-(x+h)}-\sqrt{4-x}\}\{\sqrt{4-(x+h)}+\sqrt{4-x}\}}{h\{\sqrt{4-(x+h)}+\sqrt{4-x}\}} \\
&=\lim _{h \rightarrow 0} \frac{4-(x+h)-(4-x)}{h\{\sqrt{4-(x+h)}+\sqrt{4-x}\}} \\
&=\lim _{h \rightarrow 0} \frac{-h}{h\{\sqrt{4-x-h}+\sqrt{4-x}\}}=\frac{-1}{2 \sqrt{4-x}}
\end{aligned}
\)
Example 4: Using the first principle, prove that
\(
\frac{d}{d x}\left(\frac{1}{f(x)}\right)=\frac{-f^{\prime}(x)}{[f(x)]^2}
\)
Solution: Let \(\phi=\frac{1}{f(x)}\), then \(\phi(x+h)=\frac{1}{f(x+h)}\)
\(
\therefore \frac{d}{d x}(\phi(x))=\lim _{h \rightarrow 0} \frac{\dot{\phi}(x+h)-\phi(x)}{h}
\)
\(
\begin{aligned}
&\begin{aligned}
& =\lim _{h \rightarrow 0} \frac{\frac{1}{f(x+h)}-\frac{1}{f(x)}}{h} \\
& =\lim _{h \rightarrow 0} \frac{f(x)-f(x+h)}{h f(x) f(x+h)} \\
& =\lim _{h \rightarrow 0} \frac{f(x)-f(x+h)}{h} \lim _{h \rightarrow 0} \frac{1}{f(x) f(x+h)} \\
& =-f^{\prime}(x) \frac{1}{f(x) f(x)}
\end{aligned}\\
&[f(x) \text { is differentiable } \Rightarrow f(x) \text { is continuous }\\
&\begin{aligned}
& \left.\Rightarrow \lim _{h \rightarrow 0} f(x+h)=f(x)\right] \\
& =\frac{-f^{\prime}(x)}{\{f(x)\}^2}
\end{aligned}
\end{aligned}
\)
STANDARD DERIVATIVES
Some Standard Substitutions
\(
\begin{array}{ll}
\text { Expression } & \text { Substitution } \\
\sqrt{a^2-x^2} & x=a \sin \theta \text { or } a \cos \theta \\
\sqrt{a^2+x^2} & x=a \tan \theta \text { or } a \cdot \cot \theta
\end{array}
\)
\(
\begin{array}{ll}
\sqrt{x^2-a^2} & \quad x=a \sec \theta \text { or } a \operatorname{cosec} \theta \\
\sqrt{\frac{a+x}{a-x}} \text { or } \sqrt{\frac{a-x}{a+x}} & \quad x=a \cos \theta \text { or } a \cos 2 \theta
\end{array}
\)
Example 5: If \(y=\left(1+x^{1 / 4}\right)\left(1+x^{1 / 2}\right)\left(1-x^{1 / 4}\right)\), then find \(\frac{d y}{d x}\).
Solution:
\(
\begin{aligned}
y & =\left(1+x^{1 / 4}\right)\left(1+x^{1 / 2}\right)\left(1-x^{1 / 4}\right) \\
& =\left(1+x^{1 / 4}\right)\left(1-x^{1 / 4}\right)\left(1+x^{1 / 2}\right) \\
& =\left(1-x^{1 / 2}\right)\left(1+x^{1 / 2}\right) \\
& =1-x \\
\Rightarrow & \frac{d y}{d x}=-1
\end{aligned}
\)
Example 6: If \(f(x)=x|x|\), then prove that \(f^{\prime}(x)=2|x|\).
Solution:
\(
\begin{aligned}
& f(x)=\left\{\begin{array}{l}
-x^2, x<0 \\
x^2, x \geq 0
\end{array}\right. \\
& \Rightarrow f^{\prime}(x)=\left\{\begin{array}{l}
-2 x, x<0 \\
2 x, x>0
\end{array}\right. \\
& \therefore f^{\prime}(x)=2|x|
\end{aligned}
\)
Example 7: If \(y=\sqrt{\frac{1-\cos 2 x}{1+\cos 2 x}}, x \in\left(0, \frac{\pi}{2}\right) \cup\left(\frac{\pi}{2}, \pi\right)\), then find \(\frac{d y}{d x}\).
Solution: We have
\(
\begin{aligned}
& y=\sqrt{\frac{1-\cos 2 x}{1+\cos 2 x}}=\sqrt{\frac{2 \sin ^2 x}{2 \cos ^2 x}}=\sqrt{\tan ^2 x} \\
\Rightarrow & y=|\tan x|, \text { where } x \in\left(0, \frac{\pi}{2}\right) \cup\left(\frac{\pi}{2}, \pi\right) \\
\Rightarrow & y= \begin{cases}\tan x, & x \in\left(0, \frac{\pi}{2}\right) \\
-\tan x, & x \in\left(\frac{\pi}{2}, \pi\right)\end{cases} \\
\Rightarrow & \frac{d y}{d x}= \begin{cases}\sec ^2 x, & \text { if } x \in\left(0, \frac{\pi}{2}\right) \\
-\sec ^2 x, & \text { if } x \in\left(\frac{\pi}{2}, \pi\right)\end{cases}
\end{aligned}
\)
Example 8: If \(y=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots+\frac{x^n}{n!}\), then show that \(\frac{d y}{d x}-y+\frac{x^n}{n!}=0\).
Solution:
\(
\begin{aligned}
\frac{d y}{d x} & =0+\frac{1}{1!}+\frac{1}{2!}(2 x)+\frac{1}{3!}\left(3 x^2\right)+\cdots+\frac{1}{n!}\left(n x^{n-1}\right) \\
& =1+\frac{x}{1!}+\frac{x^2}{2!}+\cdots+\frac{x^{n-1}}{(n-1)!} \\
& =\left\{1+\frac{x}{1!}+\frac{x^2}{2!}+\cdots+\frac{x^{n-1}}{(n-1)!}+\frac{x^n}{n!}\right\}-\frac{x^n}{n!}
\end{aligned}
\)
\(
\begin{aligned}
& =y-\frac{x^n}{n!} \\
& \Rightarrow \frac{d y}{d x}-y+\frac{x^n}{n!}=0
\end{aligned}
\)
DIFFERENTIATION OF INVERSE TRIGONOMETRIC FUNCTIONS
Example 9: Find \(\frac{d y}{d x}\) for \(y=\sin ^{-1}(\cos x)\), where \(x \in(0,2 \pi)\).
Solution: We have
\(
\sin ^{-1}(\cos x)=\frac{\pi}{2}-\cos ^{-1}(\cos x)
\)
\(
\begin{aligned}
& = \begin{cases}\frac{\pi}{2}-x, & \text { if } 0<x \leq \pi \\
\frac{\pi}{2}-(2 \pi-x), & \text { if } \pi<x<2 \pi\end{cases} \\
& = \begin{cases}\frac{\pi}{2}-x, & \text { if } 0<x \leq \pi \\
x-\frac{3 \pi}{2}, & \text { if } \pi<x<2 \pi\end{cases}
\end{aligned}
\)
Clearly, it is not differentiable at \(x=\pi\). Therefore,
\(
\frac{d}{d x}\left\{\sin ^{-1}(\cos x)\right\}=\left\{\begin{array}{cc}
-1, & \text { if } 0<x<\pi \\
1, & \text { if } \pi<x<2 \pi
\end{array}\right.
\)
Example 10: Differentiate \(\sin ^{-1}\left(2 x \sqrt{1-x^2}\right)\) with respect to \(x\), if
a. \(-\frac{1}{\sqrt{2}}<x<\frac{1}{\sqrt{2}}\)
b. \(\frac{1}{\sqrt{2}}<x<1\)
c. \(-1<x<-\frac{1}{\sqrt{2}}\).
Solution: Let \(y=\sin ^{-1}\left(2 x \sqrt{1-x^2}\right)\)
Substituting \(x=-\sin \theta\), where \(\theta=\sin ^{-1} x\), and \(\theta \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\), we get \(y=\sin ^{-1}(2 \sin \theta \cos \theta)=\sin ^{-1}(\sin 2 \theta)\)
(a) \(-\frac{1}{\sqrt{2}}<x<\frac{1}{\sqrt{2}} \Rightarrow-\frac{\pi}{4}<\theta<\frac{\pi}{4} \Rightarrow-\frac{\pi}{2}<2 \theta<\frac{\pi}{2}\)
\(\Rightarrow y=\sin ^{-1}(\sin 2 \theta)=2 \theta=2 \sin ^{-1} x\)
\(\Rightarrow \frac{d y}{d x}=\frac{2}{\sqrt{1-x^2}}\)
(b) \(\frac{1}{\sqrt{2}}<x<1 \Rightarrow \frac{\pi}{4}<\theta<\frac{\pi}{2} \Rightarrow \frac{\pi}{2}<2 \theta<\pi\)
\(\Rightarrow y=\sin ^{-1}(\sin 2 \theta)=\sin ^{-1}(\sin (\pi-2 \theta))=\pi-2 \theta\)
\(\Rightarrow y=\pi-2 \sin ^{-1} x\)
\(\Rightarrow \frac{d y}{d x}=0-\frac{2}{\sqrt{1-x^2}}=-\frac{2}{\sqrt{1-x^2}}\)
(c)
\(\begin{aligned}-1 & <x<-\frac{1}{\sqrt{2}} \Rightarrow-\frac{\pi}{2}<\theta<-\frac{\pi}{4} \Rightarrow-\pi<2 \theta<-\frac{\pi}{2} \\ \Rightarrow y= & \sin ^{-1}(\sin 2 \theta)=\sin ^{-1}(-\sin (\pi+2 \theta)) \\ & =\sin ^{-1}(\sin (-\pi-2 \theta)) \\ & =-\pi-2 \theta \\ \Rightarrow y & =-\pi-2 \sin ^{-1} x \\ \Rightarrow \frac{d y}{d x} & =-0-\frac{2}{\sqrt{1-x^2}}=-\frac{2}{\sqrt{1-x^2}}\end{aligned}\)
Example 11: Find \(\frac{d y}{d x}\) for \(y=\tan ^{-1}\left\{\frac{1-\cos x}{\sin x}\right\},-\pi<x<\pi\).
Solution:
\(
\begin{aligned}
y & =\tan ^{-1}\left\{\frac{1-\cos x}{\sin x}\right\}=\tan ^{-1}\left\{\frac{2 \sin ^2 \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2}}\right\} \\
& =\tan ^{-1}\left(\tan \frac{x}{2}\right)=\frac{x}{2}\left(\because-\pi<x<\pi \Rightarrow-\frac{\pi}{2}<\frac{x}{2}<\frac{\pi}{2}\right) \\
& \Rightarrow \frac{d y}{d x}=\frac{1}{2}
\end{aligned}
\)
Example 12: If \(y=\sin ^{-1}\left[x \sqrt{1-x}-\sqrt{x} \sqrt{1-x^2}\right]\), and \(0<x<1\), then find \(\frac{d y}{d x}\).
Solution:
\(
\begin{aligned}
&\begin{aligned}
y & =\sin ^{-1}\left[x \sqrt{1-x}-\sqrt{x} \sqrt{1-x^2}\right], \text { where } 0<x<1 \\
& =\sin ^{-1}\left[x \sqrt{1-(\sqrt{x})^2}-\sqrt{x} \sqrt{1-x^2}\right] \\
& =\sin ^{-1} x-\sin ^{-1} \sqrt{x} \\
& {\left[\because \sin ^{-1} x-\sin ^{-1} y=\sin ^{-1}\left(x \sqrt{1-y^2}-y \sqrt{1-x^2}\right)\right] }
\end{aligned}\\
&\text { Differentiating w.r.t. } x \text {, we get }\\
&\begin{aligned}
\frac{d y}{d x} & =\frac{1}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-(\sqrt{x})^2}} \frac{d}{d x}(\sqrt{x}) \\
& =\frac{1}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-x}} \times \frac{1}{2 \sqrt{x}}
\end{aligned}
\end{aligned}
\)
Example 13: Find \(\frac{d y}{d x}\) for \(y=\tan ^{-1} \sqrt{\frac{a-x}{a+x}},-a<x<a\).
Solution:
\(
\begin{aligned}
&y=\tan ^{-1}\left\{\sqrt{\frac{a-x}{a+x}}\right\}, \text { where }-a<x<a\\
&\text { Substituting } x=a \cos \theta \text {, we get }\\
&\begin{aligned}
y & =\tan ^{-1}\left\{\sqrt{\frac{a-a \cos \theta}{a+a \cos \theta}}\right\} \\
& =\tan ^{-1}\left\{\sqrt{\frac{1-\cos \theta}{1+\cos \theta}}\right\} \\
& =\tan ^{-1}\left\{\sqrt{\tan ^2 \frac{\theta}{2}}\right\} \\
& =\tan ^{-1}\left|\tan \frac{\theta}{2}\right|
\end{aligned}
\end{aligned}
\)
\(
\begin{gathered}
\text { Also for }-a<x<a,-1<\cos \theta<1 \\
\Rightarrow \theta \in(0, \pi) \Rightarrow \frac{\theta}{2} \in\left(0, \frac{\pi}{2}\right) \\
\therefore y=\tan ^{-1}\left|\tan \frac{\theta}{2}\right|=\tan ^{-1}\left(\tan \frac{\theta}{2}\right)=\frac{\theta}{2}=\frac{1}{2} \cos ^{-1}\left(\frac{x}{a}\right) \\
\Rightarrow \frac{d y}{d x}=-\frac{1}{2} \times \frac{1}{\sqrt{1-\frac{x^2}{a^2}}} \frac{d}{d x}\left(\frac{x}{a}\right)=-\frac{1}{2 \sqrt{a^2-x^2}}
\end{gathered}
\)
THEOREMS ON DERIVATIVES
Example 14: Find \(\frac{d y}{d x}\) for \(y=x \sin x \log x\).
Solution: We have
\(
\begin{aligned}
\frac{d}{d x}(x \sin x \log x)= & \left\{\frac{d}{d x}(x)\right\} \sin x \log x \\
& +x \frac{d}{d x}(\sin x) \log x+x \sin x \frac{d}{d x}(\log x) \\
= & 1 \times \sin x \times \log x+x \times \cos x \times \log x+x \times \sin x \times \frac{1}{x} \\
= & \sin x \log x+x \cos x \log x+\sin x
\end{aligned}
\)
Example 15: If \(y=\sqrt{\frac{1-x}{1+x}}\), prove that \(\left(1-x^2\right) \frac{d y}{d x}+y=0\).
Solution: We have
\(
y=\sqrt{\frac{1-x}{1+x}}
\)
Differentiating w. r. t. \(x\), we get
\(
\begin{aligned}
\frac{d y}{d x} & =\frac{1}{2}\left(\frac{1-x}{1+x}\right)^{(1 / 2)-1} \frac{d}{d x}\left(\frac{1-x}{1+x}\right) \\
& =\frac{1}{2} \sqrt{\frac{1+x}{1-x}} \frac{(1+x) \frac{d}{d x}(1-x)-(1-x) \frac{d}{d x}(1+x)}{(1+x)^2} \\
& =\frac{1}{2} \sqrt{\frac{1+x}{1-x}} \frac{(1+x)(-1)-(1-x)(1)}{(1+x)^2} \\
& =-\sqrt{\frac{1+x}{1-x}} \frac{1}{(1+x)^2}
\end{aligned}
\)
\(
\begin{aligned}
& \Rightarrow\left(1-x^2\right) \frac{d y}{d x}=-\sqrt{\frac{1+x}{1-x}} \frac{1}{(1+x)^2}\left(1-x^2\right) \\
& \Rightarrow\left(1-x^2\right) \frac{d y}{d x}=-\sqrt{\frac{1-x}{1+x}} \\
& \Rightarrow\left(1-x^2\right) \frac{d y}{d x}=-y \\
& \Rightarrow\left(1-x^2\right) \frac{d y}{d x}+y=0
\end{aligned}
\)
Example 16: Find the sum of the series \(1+2 x+3 x^2 +(n-1) x^{n-2}\) using differentiation.
Solution: We know that \(1+x+x^2+\cdots+x^{n-1}=\frac{1-x^n}{1-x}\)
Differentiating both sides w.r.t \(x\), we get
\(
0+1+2 x+3 x^2+\cdots+(n-1) x^{n-2}
\)
\(=\frac{(1-x) \frac{d}{d x}\left(1-x^n\right)-\left(1-x^n\right) \frac{d}{d x}(1-x)}{(1-x)^2}\)
\(
\begin{aligned}
& \Rightarrow 1+2 x+3 x^2+\cdots+(n-1) x^{n-2}=\frac{-(1-x) n x^{n-1}+\left(1-x^n\right)}{(1-x)^2} \\
& \Rightarrow 1+2 x+3 x^2+\cdots+(n-1) x^{n-2}=\frac{-n x^{n-1}+(n-1) x^n+1}{(1-x)^2}
\end{aligned}
\)
Example 17: If \(\sin y=x \cos (a+y)\), show that \(\frac{d y}{d x}=\frac{\cos ^2(a+y)}{\cos a}\), and find the value of \(d y / d x\) at \(x=0\).
Solution: We have
\(
x=\sin y / \cos (a+y) \dots(1)
\)
Differentiating w.r.t. \(y\), we get
\(
\begin{aligned}
& \Rightarrow \frac{d x}{d y}=\frac{\cos y \cos (a+y)+\sin y \sin (a+y)}{\cos ^2(a+y)} \\
& \Rightarrow \frac{d x}{d y}=\frac{\cos (a+y-y)}{\cos ^2(a+y)} \\
& \Rightarrow \frac{d y}{d x}=\frac{\cos ^2(a+y)}{\cos a}
\end{aligned}
\)
\(
\begin{aligned}
& \text { Putting } x=0 \text { in equation (1), } \sin y=0 \Rightarrow y=n \pi, \forall n \in I \\
& \therefore\left[\frac{d y}{d x}\right]_{x=0}=\frac{\cos ^2(a+n \pi)}{\cos a}=\frac{\cos ^2 a}{\cos a}=\cos a
\end{aligned}
\)
DIFFERENTIATION OF COMPOSITE FUNCTIONS (CHAIN RULE)
If \(f(x)\) and \(g(x)\) are two differentiable functions, then \(f o g\) is also differentiable, and \((f o g)^{\prime}(x)=f^{\prime}(g(x)) \cdot g^{\prime}(x)\)
Alternatively, if \(y\) is a function of \(t\), and \(t\) is a function of \(x\), then
\(
\frac{d y}{d x}=\frac{d y}{d t} \times \frac{d t}{d x}
\)
The Chain Rule:
\(
\begin{aligned}
&\text { Thus, if } y=f(t) \text { and } t=\phi(x) \text {, then } \frac{d y}{d x}=f^{\prime}(t) \text {, and }\\
&\begin{aligned}
\frac{d t}{d x} & =\phi^{\prime}(x) \\
\therefore \frac{d y}{d x} & =\frac{d y}{d t} \times \frac{d t}{d x}=f^{\prime}(t) \phi^{\prime}(x) . \text { This rule is called Chain Rule. }
\end{aligned}
\end{aligned}
\)
For example, find \(\frac{d y}{d x}\) for \(y=\sin \left(x^2+1\right)\)
Let \(y=\sin \left(x^2+1\right)\).
Putting \(u=x^2+1\), we get \(y=\sin u\)
\(
\therefore \frac{d y}{d u}=\cos u \text { and } \frac{d u}{d x}=2 x
\)
Now, \(\frac{d y}{d x}=\frac{d y}{d u} \times \frac{d u}{d x}\)
\(
\Rightarrow \frac{d y}{d x}=\cos u 2 x=2 x \cos \left(x^2+1\right)
\)
This chain rule can be extended as follows:
Let \(y=f(t), t=\phi(z), z=\psi(x)\),
then \(\frac{d y}{d x}=\frac{d y}{d t} \times \frac{d t}{d z} \times \frac{d z}{d x}=f^{\prime}(t) \phi^{\prime}(z) \psi^{\prime}(x)\)
For example, Let \(y=\log \sin x^3=\log t\)
Putting \(t=\sin x^3=\sin z\), and putting \(z=x^3\)
\(
\begin{aligned}
\Rightarrow \frac{d y}{d x}=\frac{d y}{d t} \times \frac{d t}{d z} \times \frac{d z}{d x} & =(1 / t) \cos z 3 x^2 \\
& =\left(1 / \sin x^3\right)\left(\cos x^3\right) \times 3 x^2=3 x^2 \cot x^3
\end{aligned}
\)
Example 18: If \(y=\sqrt{\log \left\{\sin \left(\frac{x^2}{3}-1\right)\right\}}\), then find \(\frac{d y}{d x}\).
Solution:
\(
\begin{aligned}
&y=\sqrt{\log \left\{\sin \left(\frac{x^2}{3}-1\right)\right\}}\\
&\text { Putting } \frac{x^2}{3}-1=v \text {, we get } \sin \left(\frac{x^2}{3}-1\right)=\sin v=u \text {, and }
\end{aligned}
\)
\(
\log \left\{\sin \left(\frac{x^2}{3}-1\right)\right\}=\log u=z
\)
we get \(y=\sqrt{z}, z=\log u, u=\sin v\) and \(v=\frac{x^2}{3}-1\)
\(
\therefore \quad \frac{d y}{d z}=\frac{1}{2 \sqrt{z}}, \frac{d z}{d u}=\frac{1}{u}, \frac{d u}{d v}=\cos v \text { and } \frac{d v}{d x}=\frac{2 x}{3}
\)
Now, \(\frac{d y}{d x}=\frac{d y}{d z} \times \frac{d z}{d u} \times \frac{d u}{d v} \times \frac{d v}{d x}\)
\(
\Rightarrow \quad \frac{d y}{d x}=\left(\frac{1}{2 \sqrt{z}}\right)\left(\frac{1}{u}\right)(\cos v)\left(\frac{2 x}{3}\right)=\frac{x}{3} \cdot \frac{\cos v}{u \sqrt{\log u}}
\)
\(
=\frac{x \cot \left(\frac{x^2}{3}-1\right)}{3 \sqrt{\log \left\{\sin \left(\frac{x^2}{3}-1\right)\right\}}}
\)
Example 19: Find \(\frac{d y}{d x}\) for \(y=\log \left(x+\sqrt{a^2+x^2}\right)\).
Solution:
\(
\begin{aligned}
&y=\log \left(x+\sqrt{a^2+x^2}\right)\\
&\text { Then } \frac{d y}{d x}=\frac{d}{d x}\left\{\log \left(x+\sqrt{a^2+x^2}\right)\right\}
\end{aligned}
\)
\(
=\frac{1}{x+\sqrt{a^2+x^2}} \frac{d}{d x}\left(x+\sqrt{a^2+x^2}\right)
\)
\(
=\frac{1}{x+\sqrt{a^2+x^2}}\left\{1+\frac{1}{2}\left(a^2+x^2\right)^{-1 / 2} \times\right. \left.\frac{d}{d x}\left(a^2+x^2\right)\right\}
\)
\(
\begin{aligned}
& =\frac{1}{x+\sqrt{a^2+x^2}}\left\{1+\frac{1}{2 \sqrt{a^2+x^2}} \times 2 x\right\} \\
& =\frac{1}{x+\sqrt{a^2+x^2}} \times \frac{\sqrt{a^2+x^2}+x}{\sqrt{a^2+x^2}} \\
& =\frac{1}{\sqrt{a^2+x^2}}
\end{aligned}
\)
DIFFERENTIATION OF IMPLICIT FUNCTIONS
If variables \(x\) and \(y\) are connected by a relation of the form \(f(x, y)=0\) and it is not possible or convenient to express \(y\) as a function \(x\), i.e., in the form \(y=\phi(x)\), then \(y\) is said to be an implicit function of \(x\). To find \(\frac{d y}{d x}\) in such a case, we differentiate both sides of the given relation with respect to \(x\), keeping in mind that the derivative of \(\phi(y)\) with respect to \(x\) is \(\frac{d \phi}{d y} \times \frac{d y}{d x}\).
For example,
\(
\frac{d}{d x}(\sin y)=\cos y \frac{d y}{d x}, \frac{d}{d x}\left(y^2\right)=2 y \frac{d y}{d x}
\)
It should be noted that \(\frac{d}{d y}(\sin y)=\cos y\)
\(
\text { but } \frac{d}{d x}(\sin y)=\cos y \frac{d y}{d x}
\)
Similarly, we have \(\frac{d}{d y}\left(y^3\right)=3 y^2\),
whereas \(\frac{d}{d x}\left(y^3\right)=3 y^2 \frac{d y}{d x}\)
A direct formula for implicit functions
Let \(f(x, y)=0\). Take all the terms towards left side and put the left side equal to \(f(x, y)\).
Then \(\frac{d y}{d x}=-\frac{\text { differentiation of } f \text { w.r.t. } x \text { keeping } y \text { as constant }}{\text { differentiation of } f \text { w.r.t. } y \text { keeping } x \text { as constant }}\)
Example 20: If \(x^2+2 x y+y^3=4\), find \(\frac{d y}{d x}\).
Solution: We have
\(
x^2+2 x y+y^3=4
\)
Differentiating both sides w.r.t. \(x\), we get
\(
\begin{aligned}
& \frac{d}{d x}\left(x^2\right)+2 \frac{d}{d x}(x y)+\frac{d}{d x}\left(y^3\right)=\frac{d}{d x}(4) \\
& \Rightarrow 2 x+2\left(x \frac{d y}{d x}+y \cdot 1\right)+3 y^2 \frac{d y}{d x}=0 \\
& \Rightarrow \frac{d y}{d x}=-\frac{2(x+y)}{\left(2 x+3 y^2\right)}
\end{aligned}
\)
Alternative method
\(Example 21:
\(
\begin{aligned}
&\text { If } y=x+\frac{1}{x+\frac{1}{x+\frac{1}{x+\cdots}}}\\
&\text { prove that } \frac{d y}{d x}=\frac{y}{2 y-x} \text {. }
\end{aligned}
\)
Solution: We have
\(
y=x+\frac{1}{x+\frac{1}{x+\frac{1}{x+\cdots}}}
\)
\(
\begin{aligned}
& \Rightarrow y=x+\frac{1}{y} \\
& \Rightarrow y^2=x y+1
\end{aligned}
\)
\(
\Rightarrow 2 y \frac{d y}{d x}=y+x \frac{d y}{d x}+0 \quad[\text { Differentiating both sides w.r.t. } x]
\)
\(
\begin{aligned}
& \Rightarrow \frac{d y}{d x}(2 y-x)=y \\
& \Rightarrow \frac{d y}{d x}=\frac{y}{2 y-x}
\end{aligned}
\)
Example 22: If \(\sqrt{x}+\sqrt{y}=4\), then find \(\frac{d x}{d y}\) at \(y=1\).
Solution: Differentiating both sides of the given equation w.r.t. \(y\), we get
\(
\begin{aligned}
& \frac{1}{2 \sqrt{x}} \frac{d x}{d y}+\frac{1}{2 \sqrt{y}}=0 \\
& \Rightarrow \frac{d x}{d y}=-\frac{\sqrt{x}}{\sqrt{y}}=\frac{\sqrt{y}-4}{\sqrt{y}} \\
& \Rightarrow\left[\frac{d x}{d y}\right]_{y=1}=\frac{1-4}{1}=-3
\end{aligned}
\)
Example 23: If \(y=\sqrt{x \log _e x}\), then find \(\frac{d y}{d x}\) at \(x=e\).
Solution:
\(
\begin{aligned}
& \frac{d y}{d x}=\frac{1}{2 \sqrt{x \log _e x}} \frac{d}{d x}\left[x \log _e x\right] \\
&=\frac{1}{2 \sqrt{x \log _e x}}\left[x \times \frac{1}{x}+1 \times \log _e x\right] \\
& \Rightarrow\left[\frac{d y}{d x}\right]_{x=e}=\frac{1}{2 \sqrt{e \times 1}}(1+1)=\frac{1}{\sqrt{e}}\left(\because \log _e e=1\right)
\end{aligned}
\)
DIFFERENTIATION OF FUNCTIONS IN PARAMETRIC FORM
Sometimes, \(x\) and \(y\) are given as functions of a single variable, i.e., \(x=\phi(t), y=\psi(t)\) are two functions and \(t\) is a variable. In such a case \(x\) and \(y\) are called parametric functions or parametric equations and \(t\) is called the parameter. To find \(\frac{d y}{d x}\) in case of parametric functions, we first obtain the relationship between \(x\) and \(y\) by eliminating the parameter \(t\) and then we differentiate it with respect to \(x\). But every time it is not convenient to eliminate the parameter. Therefore, \(\frac{d y}{d x}\) can also be obtained by the following formula
\(
\frac{d y}{d x}=\frac{d y / d t}{d x / d t}
\)
Example 24: Find \(\frac{d y}{d x}\) if \(x=a(\theta-\sin \theta)\) and \(y=a(1-\cos \theta)\).
Solution: We have, \(x=a(\theta-\sin \theta)\) and \(y=a(1-\cos \theta)\)
\(
\begin{aligned}
& \Rightarrow \frac{d x}{d \theta}=a(1-\cos \theta) \text { and } \frac{d y}{d \theta}=a \sin \theta \\
& \Rightarrow \frac{d y}{d x}=\frac{d y / d \theta}{d x / d \theta} \\
& =\frac{a \sin \theta}{a(1-\cos \theta)}=\frac{2 \sin (\theta / 2) \cos (\theta / 2)}{2 \sin ^2(\theta / 2)}=\cot \frac{\theta}{2}
\end{aligned}
\)
Example 25: If \(x=a \sec ^3 \theta\) and \(y=a \tan ^3 \theta\), find \(\frac{d y}{d x}\) at \(\theta=\frac{\pi}{3}\).
Solution: We have \(x=a \sec ^3 \theta\) and \(y=a \tan ^3 \theta\)
\(
\begin{aligned}
& \frac{d x}{d \theta}=3 a \sec ^2 \theta \frac{d}{d \theta}(\sec \theta)=3 a \sec ^3 \theta \tan \theta \\
& \text { and } \frac{d y}{d \theta}=3 a \tan ^2 \theta \frac{d}{d \theta}(\tan \theta)=3 a \tan ^2 \theta \sec ^2 \theta \\
& \Rightarrow \frac{d y}{d x}=\frac{d y / d \theta}{d x / d \theta}=\frac{3 a \tan ^2 \theta \sec ^2 \theta}{3 a \sec ^3 \theta \tan \theta}=\frac{\tan \theta}{\sec \theta}=\sin \theta \\
& \Rightarrow\left(\frac{d y}{d x}\right)_{\theta=\pi / 3}=\sin \frac{\pi}{3}=\frac{\sqrt{3}}{2}
\end{aligned}
\)
Example 26: Let \(y=x^3-8 x+7\) and \(x=f(t)\). If \(\frac{d y}{d t}=2\) and \(x=3\) at \(t=0\), then find the value of \(\frac{d x}{d t}\) at \(t=0\).
Solution: We have \(y=x^3-8 x+7\)
\(
\Rightarrow \frac{d y}{d x}=3 x^2-8
\)
It is given that when \(t=0, x=3\).
\(
\therefore \text { when } t=0, \frac{d y}{d x}=3.3^2<8=19 \text {. }
\)
Also, \(\frac{d y}{d x}=\frac{d y / d t}{d x / d t} \dots(1)\)
Since, when \(t=0, \frac{d y}{d x}=19\) and \(\frac{d y}{d t}=2\),
\(
\begin{aligned}
& \therefore \text { from }(1), 19=\frac{2}{d x / d t} \\
& \Rightarrow \frac{d x}{d t}=\frac{2}{19}
\end{aligned}
\)
DIFFERENTIATION USING LOGARITHM
\(
\begin{aligned}
& \text { If } y=\left[f_1(x)\right]^{f_2(x)} \text { or } y=f_1(x) f_2(x) f_3(x) \ldots \\
& \text { or } y=\frac{f_1(x) f_2(x) f_3(x) \ldots}{g_1(x) g_2(x) g_3(x) \ldots}
\end{aligned}
\)
then it is convenient to take the logarithm of the function first and then differentiate:
Note: Write \(y=[f(x)]^{g(x)}=e^{g(x) ln {f(x)}}\) and differentiate easily or if \(y=[f(x)]^{g(x)}\) then \(\frac{d y}{d x}\) Differential of \(y\) treating \(f(x)\) as constant + Differential of \(y\) treating \(g(x)\) as constant.
Let \(y=\{f(x)\}^{g(x)}\)
Taking logarithm of both the sides, we have
\(
\log y=g(x) \cdot \log {f(x)}
\)
Differentiating w.r.t. \(x\), we get
\(
\begin{array}{ll}
& \frac{1}{y} \frac{d y}{d x}=g(x) \frac{1}{f(x)} \frac{d f(x)}{d x}+\log \{f(x)\} \frac{d g(x)}{d x} \\
\therefore & \frac{d y}{d x}=y\left\{\frac{g(x)}{f(x)} \frac{d f(x)}{d x}+\log {f(x)} \frac{d g(x)}{d x}\right\}
\end{array}
\)
\(
\frac{{dy}}{{dx}}={y}\left[{~g}({x}) \cdot \frac{{f}^{\prime}({x})}{{f}({x})}+\log ({f}({x})) {g}^{\prime}({x})\right]
\)
For example, If \(y=(\sin x)^{\log x}\) then \(\frac{d y}{d x}=\) ?
\(
\frac{d y}{d x}=(\sin x)^{\log x}\left(\frac{\log (\sin x)}{x}+\log x \cdot \cot x\right)
\)
Example 27: If \(x^m y^n=(x+y)^{m+n}\), prove that \(\frac{d y}{d x}=\frac{y}{x}\).
Solution: We have \(x^m y^n=(x+y)^{m+n}\).
Taking \(\log\) on both sides, we get
\(
m \log x+n \log y=(m+n) \log (x+y)
\)
Differentiating both sides w.r.t. \(x\), we get
\(
\begin{aligned}
& m \frac{1}{x}+n \frac{1}{y} \frac{d y}{d x}=\frac{m+n}{x+y} \frac{d}{d x}(x+y) \\
& \Rightarrow\left(\frac{m}{x}+\frac{n}{y}\right) \frac{d y}{d x}=\frac{m+n}{x+y}\left(1+\frac{d y}{d x}\right) \\
& \Rightarrow\left\{\frac{n}{y}-\frac{m+n}{x+y}\right\} \frac{d y}{d x}=\frac{m+n}{x+y}-\frac{m}{x} \\
& \Rightarrow\left\{\frac{n x+n y-m y-n y}{y(x+y)}\right\} \frac{d y}{d x}=\left\{\frac{m x+n x-m x-m y}{(x+y) x}\right\} \\
& \Rightarrow \frac{n x-m y}{y(x+y)} \frac{d y}{d x}=\frac{n x-m y}{(x+y) x} \\
& \Rightarrow \frac{d y}{d x}=\frac{y}{x}
\end{aligned}
\)
Example 28: Find \(\frac{d y}{d x}\) for \(y=(\sin x)^{\log x}\).
Solution: Let \(y=(\sin x)^{\log x}\).
Then, \(y=e^{\log x \log \sin x}\)
Differentiating both sides w.r.t. \(x\), we get
\(
\begin{aligned}
& \frac{d y}{d x}=e^{\log x \log \sin x} \frac{d}{d x}\{\log x \log \sin x\} \\
& \Rightarrow \frac{d y}{d x}=(\sin x)^{\log x} \\
& \quad \quad \times\left\{\log \sin x \frac{d}{d x}(\log x)+\log x \frac{d}{d x}(\log \sin x)\right\} \\
& \Rightarrow \frac{d y}{d x}=(\sin x)^{\log x}\left\{\frac{\log \sin x}{x}+\log x \frac{1}{\sin x} \cos x\right\} \\
& \Rightarrow \frac{d y}{d x}=(\sin x)^{\log x}\left\{\frac{\log \sin x}{x}+\cot x \log x\right\}
\end{aligned}
\)
Example 29: If \(y=x^{x^{x^{x^{-\infty}}}}\), find \(\frac{d y}{d x}\)
Solution: Since by deleting a single term from an infinite series, it remains same.
Therefore, the given function may be written as
\(
\begin{aligned}
& y=x^y \\
& \Rightarrow \log y=y \log x \\
& \Rightarrow \frac{1}{y} \frac{d y}{d x}=\frac{d y}{d x} \times \log x+y \frac{1}{x} [\text { Diff. both sides w.r.t. } x] \\
& \Rightarrow \frac{d y}{d x} \frac{\{1-y \log x\}}{y}=\frac{y}{x} \\
& \Rightarrow \frac{d y}{d x}=\frac{y^2}{x(1-y \log x)}
\end{aligned}
\)
Example 30: Find the derivative of \(\frac{\sqrt{x}(x+4)^{3 / 2}}{(4 x-3)^{4 / 3}}\) w.r.t. \(x\).
Solution: \(\text { Let } y=\frac{\sqrt{x}(x+4)^{3 / 2}}{(4 x-3)^{4 / 3}}\)
Taking \(\log\) of both sides, we get
\(
\log y=\frac{1}{2} \log x+\frac{3}{2} \log (x+4)-\frac{4}{3} \log (4 x-3)
\)
Differentiating both sides w.r.t. \(x\), we get
\(
\begin{aligned}
& \frac{1}{y} \frac{d y}{d x}=\frac{1}{2 x}+\frac{3}{2} \frac{1}{x+4}-\frac{4}{3} \times \frac{1}{4 x-3} \times 4 \\
& \Rightarrow \frac{d y}{d x}=y\left\{\frac{1}{2 x}+\frac{3}{2(x+4)}-\frac{16}{3(4 x-3)}\right\} \\
& \Rightarrow \frac{d y}{d x}=\frac{\sqrt{x}(x+4)^{3 / 2}}{(4 x-3)^{4 / 3}}\left\{\frac{1}{2 x}+\frac{3}{2(x+4)}-\frac{16}{3(4 x-3)}\right\}
\end{aligned}
\)
Example 31: If \(x<1\), prove that
\(
\frac{1}{1+x}+\frac{2 x}{1+x^2}+\frac{4 x^3}{1+x^4}+\cdots \infty=\frac{1}{1-x} .
\)
Solution: The given series is in the form
\(
\frac{f_1^{\prime}(x)}{f_1(x)}+\frac{f_2^{\prime}(x)}{f_2(x)}+\frac{f_3^{\prime}(x)}{f_3(x)}+\cdots \infty
\)
Then consider the product \(f_1(x) \times f_2(x) \times f_3(x) \ldots f_n(x)\)
Now \((1-x)(1+x)\left(1+x^2\right)\left(1+x^4\right) \ldots\left(1+x^{2^{n-1}}\right) \dots(1)\)
\(
\begin{aligned}
& =\left(1-x^2\right)\left(1+x^2\right)\left(1+x^4\right) \ldots\left(1+x^{2^{n-1}}\right) \\
& =\left(1-x^4\right)\left(1+x^4\right) \ldots\left(1+x^{2^{n-1}}\right) \\
& \vdots \\
& =\left(1-x^{2^{n-1}}\right)\left(1+x^{2^{n-1}}\right) \\
& =1-x^{2^n}
\end{aligned}
\)
Now when \(n \rightarrow \infty, x^{2^{n-1}} \rightarrow 0(\because x<1)\)
\(\therefore\) taking \(n \rightarrow \infty\), in (1), we get
\(
(1-x)(1+x)\left(1+x^2\right)\left(1+x^4\right) \ldots=1
\)
Taking logarithm, we get
\(
\log (1-x)+\log (1+x)+\log \left(1+x^2\right)+\log \left(1+x^4\right)+\ldots=0
\)
Differentiating w.r.t. ‘ \(x\) ‘, we get
\(
\begin{aligned}
& -\frac{1}{1-x}+\frac{1}{1+x}+\frac{2 x}{1+x^2}+\frac{4 x^3}{1+x^4}+\cdots=0 \\
& \text { or } \frac{1}{x+1}+\frac{2 x}{1+x^2}+\frac{4 x^3}{1+x^4}+\cdots \infty=\frac{1}{1-x}
\end{aligned}
\)
DIFFERENTIATION OF ONE FUNCTION W.R.T. OTHER FUNCTION
Let \(u=f(x)\) and \(v=g(x)\) be the two functions of \(x\). Then to find the derivative of \(f(x)\) w.r.t. \(g(x)\), i.e., to find \(\frac{d u}{d v}\), we use the formula: \(\frac{d u}{d v}=\frac{d u / d x}{d v / d x}\).
Thus, to find the derivative of \(f(x)\) w.r.t. \(g(x)\), we first differentiate both w.r.t. \(x\). and then divide the derivative of \(f(x)\) w.r.t. \(x\) by the derivative of \(g(x)\) w.r.t. \(x\).
Example 32: Differentiate \(\log \sin x\) w.r.t. \(\sqrt{\cos x}\).
Solution: Let \(u=\log \sin x\) and \(v=\sqrt{\cos x}\)
Then, \(\frac{d u}{d x}=\cot x\) and \(\frac{d v}{d x}=-\frac{\sin x}{2 \sqrt{\cos x}}\)
\(
\begin{aligned}
\Rightarrow \frac{d u}{d v}=\frac{d u / d x}{d v / d x} & =\frac{\cot x}{-\frac{\sin x}{2 \sqrt{\cos x}}} \\
& =-2 \sqrt{\cos x} \cot x \operatorname{cosec} x
\end{aligned}
\)
Example 33: Differentiate \(\tan ^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)\) w.r.t. \(\tan ^{-1} x\),
Solution: Let \(u=\tan ^{-1} \cdot\left(\frac{\sqrt{1+x^2}-1}{x}\right)\) and \(v=\tan ^{-1} x\).
Putting \(x=\tan \theta\),
we get \(u=\tan ^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)\)
\(
\begin{aligned}
& =\tan ^{-1}\left(\frac{\sec \theta-1}{\tan \theta}\right) \\
& =\tan ^{-1}\left(\frac{1-\cos \theta}{\sin \theta}\right) \\
& =\tan ^{-1}\left(\tan \frac{\theta}{2}\right)=\frac{1}{2} \theta=\frac{1}{2} \tan ^{-1} x
\end{aligned}
\)
Thus, we have \(u=\frac{1}{2} \tan ^{-1} x\) and \(v=\tan ^{-1} x\)
\(
\begin{aligned}
& \Rightarrow \frac{d u}{d x}=\frac{1}{2} \times \frac{1}{1+x^2} \text { and } \frac{d v}{d x}=\frac{1}{1+x^2} \\
& \Rightarrow \frac{d u}{d v}=\frac{d u / d x}{d v / d x}=\frac{1}{2\left(1+x^2\right)}\left(1+x^2\right)=\frac{1}{2}
\end{aligned}
\)
Example 34: Find the derivative of \(f(\tan x)\) w.r.t. \(g(\sec x)\) at \(x=\frac{\pi}{4}\), where \(f^{\prime}(1)=2\) and \(g^{\prime}(\sqrt{2})=4\).
Solution:
\(
\begin{aligned}
&\text { Let } u=f(\tan x) \text { and } v=g(\sec x)\\
&\begin{aligned}
& \Rightarrow \frac{d u}{d x}=f^{\prime}(\tan x) \sec ^2 x \\
& \text { and } \frac{d v}{d x}=g^{\prime}(\sec x) \sec x \tan x \\
& \Rightarrow \frac{d u}{d v}=\frac{d u}{d x} / \frac{d v}{d x}=\frac{f^{\prime}(\tan x) \sec ^2 x}{g^{\prime}(\sec x) \sec x \tan x} \\
& \Rightarrow\left[\frac{d u}{d v}\right]_{x=\frac{\pi}{4}}=\frac{f^{\prime}\left(\tan \frac{\pi}{4}\right)}{g^{\prime}\left(\sec \frac{\pi}{4}\right) \sin \frac{\pi}{4}}=\frac{f^{\prime}(1) \sqrt{2}}{g^{\prime}(\sqrt{2})} \\
& =\frac{2 \sqrt{2}}{4}=\frac{1}{\sqrt{2}}
\end{aligned}
\end{aligned}
\)
DIFFERENTIATION OF DETERMINANTS
To differentiate a determinant, we differentiate one row (or column) at a time, keeping others unchanged.
For example, if
\(\Delta(x)=\left|\begin{array}{ll}
f(x) & g(x) \\
u(x) & v(x)
\end{array}\right|, \text { then }
\)
\(
\frac{d}{d x}\{\Delta(x)\}=\left|\begin{array}{cc}
f^{\prime}(x) & g^{\prime}(x) \\
u(x) & v(x)
\end{array}\right|+\left|\begin{array}{cc}
f(x) & g(x) \\
u^{\prime}(x) & v^{\prime}(x)
\end{array}\right|
\)
Also,
\(
\frac{d}{d x}\{\Delta(x)\}=\left|\begin{array}{ll}
f^{\prime}(x) & g(x) \\
u^{\prime}(x) & v(x)
\end{array}\right|+\left|\begin{array}{ll}
f(x) & g^{\prime}(x) \\
u(x) & v^{\prime}(x)
\end{array}\right|
\)
Similar results hold for the differentiation of determinants of higher order. Following examples will illustrate the same.
Example 35: If \(f(x)=\left|\begin{array}{ccc}x+a^2 & a b & a c \\ a b & x+b^2 & b c \\ a c & b c & x+c^2\end{array}\right|\), then prove that \(f^{\prime}(x)=3 x^2+2 x\left(a^2+b^2+c^2\right)\).
Solution: We have
\(
f(x)=\left|\begin{array}{ccc}
x+a^2 & a b & a c \\
a b & x+b^2 & b c \\
a c & b c & x+c^2
\end{array}\right|
\)
\(
\begin{array}{r}
\Rightarrow f^{\prime}(x)=\left|\begin{array}{ccc}
1 & 0 & 0 \\
a b & x+b^2 & b c \\
a c & b c & x+c^2
\end{array}\right|+\left|\begin{array}{ccc}
x+a^2 & a b & a c \\
0 & 1 & 0 \\
a c & b c & x+c^2
\end{array}\right| \\
+\left|\begin{array}{ccc}
x+a^2 & a b & a c \\
a b & x+b^2 & b c \\
0 & 0 & 1
\end{array}\right|
\end{array}
\)
\(
\begin{aligned}
\Rightarrow f^{\prime}(x)=\left|\begin{array}{cc}
x+b^2 & b c \\
b c & x+c^2
\end{array}\right|+\left|\begin{array}{cc}
x+a^2 & a c \\
a c & x+c^2
\end{array}\right| & \\
& +\left|\begin{array}{cc}
x+a^2 & a b \\
a b & x+b^2
\end{array}\right|
\end{aligned}
\)
\(
\begin{aligned}
\Rightarrow f^{\prime}(x)= & {\left[\left(x+b^2\right)\left(x+c^2\right)-b^2 c^2\right]+\left[\left(x+a^2\right)\left(x+c^2\right)\right.} \\
& \left.-a^2 c^2\right]+\left[\left(x+a^2\right)\left(x+b^2\right)-a^2 b^2\right] \\
\Rightarrow & f^{\prime}(x)=3 x^2+2 x\left(a^2+b^2+c^2\right)
\end{aligned}
\)
Example 36: If \(f(x)=\left|\begin{array}{ccc}\sec \theta & \operatorname{tan}^2 \theta & 1 \\ \theta \sec x & \tan x & x \\ 1 & \tan x-\tan \theta & 0\end{array}\right|\), then \(f^{\prime}(\theta)\) is
Solution: We have
\(
f(x)=\left|\begin{array}{ccc}
\sec \theta & \tan ^2 \theta & 1 \\
\theta \sec x & \tan x & x \\
1 & \tan x-\tan \theta & 0
\end{array}\right|
\)
\(
f^{\prime}(x)=\left|\begin{array}{ccc}
0 & 0 & 0 \\
\theta \sec x & \tan x & x \\
1 & \tan x-\tan \theta & 0
\end{array}\right|
\)
\(
+\left|\begin{array}{ccc}
\sec \theta & \tan ^2 \theta & 1 \\
\theta \sec x \tan x & \sec ^2 x & 1 \\
1 & \tan x-\tan \theta & 0
\end{array}\right|+\left|\begin{array}{ccc}
\sec \theta & \tan ^2 \theta & 1 \\
\theta \sec x & \tan x & x \\
0 & \sec ^2 x & 0
\end{array}\right|
\)
\(
\begin{array}{ll}
\therefore & f^{\prime}(\theta)=0+\left|\begin{array}{ccc}
\sec \theta & \tan ^2 \theta & 1 \\
\theta \sec \theta \tan \theta \sec ^2 \theta & 1 \\
1 & 0 & 0
\end{array}\right|+\left|\begin{array}{ccc}
\sec \theta & \tan ^2 \theta & 1 \\
\theta \sec \theta & \tan \theta & \theta \\
0 & \sec ^2 \theta & 0
\end{array}\right| \\
\Rightarrow & f^{\prime}(\theta)=\tan ^2 \theta-\sec ^2 \theta+\sec ^2 \theta(\theta \sec \theta-\theta \sec \theta)=-1 .
\end{array}
\)
HIGHER ORDER DERIVATIVES
If \(y=y(x)\), then \(\frac{d y}{d x}\), the derivative of \(y\) with respect to \(x\), is itself, in general, a function of \(x\) and can be differentiated again. We call \(\frac{d y}{d x}\) as the first-order derivative of \(y\) with respect to \(x\) and the derivates of \(\frac{d y}{d x}\) w.r.t. \(x\) as the second-order derivative of \(y\) w.r.t. \(x\), and it is denoted by \(\frac{d^2 y}{d x^2}\). Similarly, the derivative of \(\frac{d^2 y}{d x^2}\) w.r.t. \(x\) is termed as the third-order derivative of \(y\) w.r.t. \(x\) and is denoted by \(\frac{d^3 y}{d x^3}\) and so on. The \(n\)th order derivative of \(y\) w.r.t. \(x\) is denoted by \(\frac{d^n y}{d x^n}\).
If \(y=f(x)\), then the other alternative notations for \(\frac{d y}{d x}, \frac{d^2 y}{d x^2}, \frac{d^2 y}{d x^3}, \ldots, \frac{d^n y}{d x^n}\) are
\(
\begin{aligned}
& y_1, y_2, y_3, \ldots, y_n(n) \\
& y^{\prime}, y^{\prime \prime}, y^{\prime \prime \prime}, \ldots, y^{(n)} \\
& f^{\prime}(x), f^{\prime \prime}(x), f^{\prime \prime \prime}(x), \ldots, f^n(x)
\end{aligned}
\)
The values of \(n\)th derivatives at \(x=a\) are denoted by
\(
y_n(a), y^n(a), D^n y(a), f^n(a) \text { or }\left(\frac{d^n y}{d x^n}\right)_{x=a}
\)
Example 37:
\(
\begin{aligned}
&\text { If } y=e^{\tan ^{-1} x} \text {, then prove that }\\
&\left(1+x^2\right) \frac{d^2 y}{d x^2}=(1-2 x) \frac{d y}{d x}
\end{aligned}
\)
Solution:
\(
\begin{aligned}
& y=e^{\tan ^{-1} x} \Rightarrow \frac{d y}{d x}=\frac{e^{\tan ^{-1} x}}{1+x^2} \\
& \Rightarrow \frac{d^2 y}{d x^2}=\frac{\left(1+x^2\right) \frac{e^{\tan ^{-1} x}}{\left(1+x^2\right)}-e^{\tan ^{-1} x}(2 x)}{\left(1+x^2\right)^2} \\
& \Rightarrow \frac{d^2 y}{d x^2}=\frac{(1-2 x) e^{\tan ^{-1} x}}{\left(1+x^2\right)^2} \Rightarrow \frac{d^2 y}{d x^2}\left(1+x^2\right)=(1-2 x) \frac{d y}{d x}
\end{aligned}
\)
Example 38: If \(y=\left(x^2-1\right)^m\), then the \((2 m)\) th differential coefficient of \(y\) is ____.
Solution: Expanding binomially, we get
\(
y=\left(x^2-1\right)^m={ }^m C_0 x^{2 m}+{ }^m C_1 x^{2 m-2}(-1)+\ldots
\)
So on differentiating, all the terms, except first, reduce to zero, therefore
\(
\begin{aligned}
\frac{d^{2 m}}{d x^{2 m}}\left(x^2-1\right)^m & ={ }^m \mathrm{C}_0 2 m(2 m-1)(2 m-2) \ldots 1 \\
& =(2 m)!
\end{aligned}
\)
Example 39: If \(y=x \log \{x /(a+b x)\}\), then show that
\(
x^3 \frac{d^2 y}{d x^2}=\left(x \frac{d y}{d x}-y\right)^2
\)
Solution:
\(
\begin{aligned}
& \text { Given } y / x=[\log x-\log (a+b x)] \\
& \frac{1}{x} \frac{d y}{d x}-\frac{1}{x^2} y=\frac{1}{x}-\frac{b}{a+b x} \\
& \Rightarrow x \frac{d y}{d x}-y=\frac{a x}{a+b x} \dots(1)
\end{aligned}
\)
Differentiating again w.r.t. \(x\)
\(
\begin{aligned}
& \left(x \frac{d^2 y}{d x^2}+\frac{d y}{d x}\right)-\frac{d y}{d x}=\frac{a^2}{(a+b x)^2} \\
& \therefore x^3 \frac{d^2 y}{d x^2}=\frac{a^2 x^2}{(a+b x)^2}=\left(x \frac{d y}{d x}-y\right)^2 \text { by (1) }
\end{aligned}
\)
Example 40: If \(y=(a x+b) /\left(x^2+c\right)\), then show that \(\left(2 x y^{\prime}+y\right) y^{\prime \prime \prime}=3\left(x y^{\prime \prime}+y^{\prime}\right) y^{\prime \prime}\), where \(a, b, c\) are constants and dashes denote differentiation w.r.t. \(x\).
Solution: Given that \(y\left(x^2+c\right)=a x+b\)
Differentiating w.r.t. \(x\)
\(
y^{\prime}\left(x^2+c\right)+2 x y=a \dots(1)
\)
Differentiating again w.r.t. \(x\)
\(
\begin{aligned}
& y^{\prime \prime}\left(x^2+c\right)+y^{\prime} 2 x+2 x y^{\prime}+2 y=0 \\
& \Rightarrow y^{\prime \prime}\left(x^2+c\right)+2\left(2 x y^{\prime}+y\right)=0 \dots(2)
\end{aligned}
\)
Differentiating again w.r.t. \(x\)
\(
\begin{aligned}
& y^{\prime \prime \prime}\left(x^2+c\right)+2 x y^{\prime \prime}+2\left(2 x y^{\prime \prime}+3 y^{\prime}\right)=0 \\
& \Rightarrow y^{\prime \prime \prime}\left(x^2+c\right)+6\left(x y^{\prime \prime}+y^{\prime}\right)=0 \dots(3)
\end{aligned}
\)
PROBLEMS BASED ON FIRST DEFINITION OF DERIVATIVE
Example 41: A function \(f: R \rightarrow R\) satisfies the equation \(f(x+y)=f(x) f(y)\) for all \(x, y \in R\) and \(f(x) \neq 0\) for all \(x \in R\). If \(f(x)\) is differentiable at \(x=0\) and \(f^{\prime}(0)=2\), then prove that \(f^{\prime}(x)=2 f(x)\).
Solution:
\(
\begin{aligned}
& \text { We have } f(x+y)=f(x) f(y) \text { for all } x, y \in R \\
& \therefore f(0)=f(0) f(0) \Rightarrow f(0)\{f(0)-1\}=0 \Rightarrow f(0)=1 \\
& \qquad[\because f(0) \neq 0]
\end{aligned}
\)
\(
\begin{aligned}
& \text { Now, } f^{\prime}(0)=2 \\
& \Rightarrow \lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}=2 \\
& \Rightarrow \lim _{h \rightarrow 0} \frac{f(h)-1}{h}=2 \quad(\because f(0)=1) \dots(1) \\
& \text { Now, } f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\
& =\lim _{h \rightarrow 0} \frac{f(x) f(h)-f(x)}{h} \quad(\because f(x+y)=f(x) f(y))
\end{aligned}
\)
\(
=f(x)\left(\lim _{h \rightarrow 0} \frac{f(h)-1}{h}\right)=2 f(x) \text { [Using (1)] }
\)
Example 42: Let \(f: R \rightarrow R\) satisfying \(|f(x)| \leq x^2, \forall x \in R\), differentiable at \(x=0\) then find \(f^{\prime}(0)\).
Solution:
\(
\begin{aligned}
& \text { Since, }|f(x)| \leq x^2, \forall x \in R \dots(1) \\
& \therefore \operatorname{At} x=0,|f(0)| \leq 0 \Rightarrow f(0)=0 \dots(2) \\
& \therefore f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{f(h)-f(0)}{h}=\lim _{h \rightarrow 0} \frac{f(h)}{h} \dots(3) \\
& \text { Now, }\left|\frac{f(h)}{h}\right| \leq|h| \text { (from(1)) } \\
& \Rightarrow-|h| \leq \frac{f(h)}{h} \leq|h|
\end{aligned}
\)
\(
\Rightarrow \lim _{h \rightarrow 0} \frac{f(h)}{h} \rightarrow 0 \quad \text { (using Sandwich Theorem) } \dots(4)
\)
\(\therefore\) from (3) and (4), we get \(f^{\prime}(0)=0\).
Example 43: Let \(f\left(\frac{x+y}{2}\right)=\frac{f(x)+f(y)}{2}\) for all real \(x\) and \(y\). If \(f^{\prime}(0)\) exists and equals -1 and \(f(0)=1\), then find \(f(2)\).
Solution: Since \(f\left(\frac{x+y}{2}\right)=\frac{f(x)+f(y)}{2}\)
Replacing \(x\) by \(2 x\) and \(y\) by 0 , then \(f(x)=\frac{f(2 x)+f(0)}{2}\)
\(
\Rightarrow f(2 x)+f(0)=2 f(x) \Rightarrow f(2 x)-2 f(x)=-f(0) \dots(1)
\)
Now, \(f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\)
\(
\begin{aligned}
& =\lim _{h \rightarrow 0} \frac{f\left(\frac{2 x+2 h}{2}\right)-f(x)}{h} \\
& =\lim _{h \rightarrow 0}\left\{\frac{\frac{f(2 x)+f(2 h)}{2}-f(x)}{h}\right\}
\end{aligned}
\)
\(
\begin{aligned}
& =\lim _{h \rightarrow 0}\left\{\frac{f(2 x)+f(2 h)-2 f(x)}{2 h}\right\} \\
& =\lim _{h \rightarrow 0}\left\{\frac{f(2 h)-f(0)}{2 h}\right\} (\text { from }(1)) \\
& =f^{\prime}(0) \\
& =-1 \quad \forall x \in R \text { (given) }
\end{aligned}
\)
Integrating, we get \(f(x)=-x+c\)
Putting \(x=0\), then \(f(0)=0+c=1 \text { (given) }\)
\(
\therefore c=1
\)
then \(f(x)=1-x\)
\(
\therefore f(2)=1-2=-1
\)
Alternate Method:
\(
\because f\left(\frac{x+y}{2}\right)=\frac{f(x)+f(y)}{2}
\)
Differentiating both sides w.r.t. \(x\) treating \(y\) as constant.
\(
\therefore f^{\prime}\left(\frac{x+y}{2}\right) \cdot \frac{1}{2}=\frac{f^{\prime}(x)+0}{2} \Rightarrow f^{\prime}\left(\frac{x+y}{2}\right)=f^{\prime}(x)
\)
Replacing \(x\) by 0 and \(y\) by \(2 x\),
\(
\text { then } f^{\prime}(x)=f^{\prime}(0)=-1 \text { (given) }
\)
Integrating, we have \(f(x)=-x+c\).
\(
\text { Putting } x=0, f(0)=0+c=1 \text { (given) }
\)
\(
\therefore c=1
\)
Hence, \(f(x)=-x+1\)
\(
\text { then } f(2)=-2+1=-1 \text {. }
\)
Example 44: If \(f(x)+f(y)=f\left(\frac{x+y}{1-x y}\right)\) for all \(x, y \in R(x y \neq 1)\) and \(\lim _{x \rightarrow 0} \frac{f(x)}{x}=2\). Find \(f\left(\frac{1}{\sqrt{3}}\right)\) and \(f^{\prime}(1)\).
Solution:
\(
\begin{aligned}
& f(x)+f(y)=f\left(\frac{x+y}{1-x y}\right) \dots(1) \\
& \text { Putting } x=y=0, \text { we get } f(0)=0 \\
& \text { Putting } y=-x, \text { we get } f(+x)+f(-x)=f(0) \\
& \Rightarrow f(-x)=-f(x) \dots(2) \\
& \text { also, } \lim _{x \rightarrow 0} \frac{f(x)}{x}=2 \\
& \text { Now, } f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \dots(3)
\end{aligned}
\)
\(
=\lim _{h \rightarrow 0} \frac{f(x+h)+f(-x)}{h} \quad(\text { using }(2)-f(x)=f(-x))
\)
\(
f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f\left(\frac{x+h-x}{1-(x+h)(-x)}\right)}{h} \text { (using (1)) }
\)
\(
\begin{aligned}
& \Rightarrow f^{\prime}(x)=\lim _{h \rightarrow 0}\left[\frac{f\left(\frac{h}{1+x(x+h)}\right)}{h}\right] \\
& \Rightarrow f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f\left(\frac{h}{1+x h+x^2}\right)}{\left(\frac{h}{1+x h+x^2}\right)} \times\left(\frac{1}{1+x h+x^2}\right) \\
& \Rightarrow f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f\left(\frac{h}{1+x h+x^2}\right)}{\left(\frac{h}{1+x h+x^2}\right)} \times \lim _{h \rightarrow 0} \frac{1}{1+x h+x^2} \quad \left(\text { using } \lim _{x \rightarrow 0} \frac{f(x)}{x}=2\right)
\end{aligned}
\)
\(
\Rightarrow f^{\prime}(x)=2 \times \frac{1}{1+x^2} \quad \Rightarrow f^{\prime}(x)=\frac{2}{1+x^2}
\)
Integrating both sides, we get \(f(x)=2 \tan ^{-1}(x)+c\), where \(f(0)=0 \Rightarrow c=0\) Thus, \(f(x)=2 \tan ^{-1} x\).
Hence, \(f\left(\frac{1}{\sqrt{3}}\right)=2 \tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)=2 \frac{\pi}{6}=\frac{\pi}{3}\), and
\(
f^{\prime}(1)=\frac{2}{1+1^2}=\frac{2}{2}=1
\)
Example 45: Suppose \(p(x)=a_0+a_1 x+a_2 x^2+\cdots+a_n x^n\). If \(|p(x)| \leq\left|e^{x-1}-1\right|\) for all \(x \geq 0\), prove that \(\left|a_1+2 a_2+\cdots+n a_n\right| \leq 1\).
Solution:
\(
\begin{aligned}
& \text { Given } p(x)=a_0+a_1 x+a_2 x^2+\cdots+a_n x^n \\
& \therefore p^{\prime}(x)=0+a_1+2 a_2 x+\cdots+n a_n x^{n-1} \\
& \Rightarrow p^{\prime}(1)=a_1+2 a_2+\cdots+n a_n \dots(1)
\end{aligned}
\)
\(
\text { Now, }|p(1)| \leq 0,\left(\because\left|e^{1-1}-1\right|=\left|e^0-1\right|=|1-1|=0\right)
\)
\(
\Rightarrow|p(1)| \leq 0 \Rightarrow p(1)=0 (\therefore|p(1)| \geq 0)
\)
\(
\begin{aligned}
& \text { As }|p(x)| \leq\left|e^{x-1}-1\right| \text {, we get } \\
& |p(1+h)| \leq\left|e^h-1\right| \forall h>-1, h \neq 0 \\
& \Rightarrow|p(1+h)-p(1)| \leq\left|e^h-1\right| (\because p(1)=0)
\end{aligned}
\)
\(
\begin{aligned}
&\Rightarrow\left|\frac{p(1+h)-p(1)}{h}\right| \leq\left|\frac{e^h-1}{h}\right|\\
&\text { Taking limit as } h \rightarrow 0 \text {, then }\\
&\begin{aligned}
& \Rightarrow \lim _{h \rightarrow 0}\left|\frac{p(1+h)-p(1)}{h}\right| \leq \lim _{h \rightarrow 0}\left|\frac{e^h-1}{h}\right| \\
& \Rightarrow\left|p^{\prime}(1)\right| \leq 1 \\
& \Rightarrow\left|a_1+2 a_2+\cdots+n a_n\right| \leq 1 \text { (from (1)) }
\end{aligned}
\end{aligned}
\)
MISCELLANEOUS SOLVED PROBLEMS
Q1: Derivative of \(\tan ^{-1}\left\{\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}}\right\}\) with respect to \(\cos ^{-1} x^2\) is
a. \(-1 / 2\)
b. \(1 / 2\)
c. -1
d. 1
Solution: (a)
\(
\text { Let } \begin{aligned}
u & =\tan ^{-1}\left\{\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}}\right\} \\
& =\tan ^{-1}\left\{\frac{1-\sqrt{\frac{1-x^2}{1+x^2}}}{1+\sqrt{\frac{1-x^2}{1+x^2}}}\right\}
\end{aligned}
\)
Let \(x^2=\cos 2 \theta\)
\(
\begin{aligned}
\Rightarrow u & =\tan ^{-1}\left\{\frac{1-\tan \theta}{1+\tan \theta}\right\} \\
& =\tan ^{-1}\left\{\tan \left(\frac{\pi}{4}-\theta\right)\right\}=\frac{\pi}{4}-\theta=\frac{\pi}{4}-\frac{1}{2} \cos ^{-1} x^2
\end{aligned}
\)
Also \(v=\cos ^{-1} x^2\)
\(
\Rightarrow \frac{d u}{d v}=-\frac{1}{2}
\)
Q2: Let \(f\) and \(g\) be differentiable functions satisfying \(g^{\prime}(a) =2, g(a)=b\) and \(f \circ g=I\) (identity function). Then \(f^{\prime}(b)\) is equal to
a. \(\frac{1}{2}\)
b. 2
c. \(\frac{2}{3}\)
d. None of these
Solution:
\(
\begin{aligned}
& \text { (a) Given } f o g=I \\
& \Rightarrow f o g(x)=x \text { for all } x \\
& \Rightarrow f^{\prime}(g(x)) g^{\prime}(x)=1 \text { for all } x \\
& \Rightarrow f^{\prime}(g(a))=\frac{1}{g^{\prime}(a)}=\frac{1}{2} \\
& \Rightarrow f^{\prime}(b)=\frac{1}{2} (\because g(a)=b)
\end{aligned}
\)
Q3: If \(f(x)=\cot ^{-1}\left(\frac{x^x-x^{-x}}{2}\right)\), then \(f^{\prime}(1)\) is
a. -1
b. 1
c. \(\ln 2\)
d. \(-\ln 2\)
Solution: (a) \(\frac{d}{d x}\left(x^x\right)=x^x(1+\log x), \frac{d}{d x}\left(x^{-x}\right)=-x^{-x}(1+\log x)\)
\(
\begin{aligned}
\Rightarrow f^{\prime}(x) & =\frac{d}{d x}(f(x)) \\
& =-\frac{1}{1+\left(\frac{x^x-x^{-x}}{2}\right)^2} \cdot \frac{d}{d x}\left(\frac{x^x-x^{-x}}{2}\right) \\
& =-\frac{4}{4+\left(x^x-x^{-x}\right)^2} \frac{1}{2}\left(x^x(1+\log x)+x^{-x}(1+\log x)\right)
\end{aligned}
\)
\(
=-2 \frac{\left(x^x+x^{-x}\right)(1+\log x)}{\left(x^x-x^{-x}\right)^2}
\)
\(
\Rightarrow f^{\prime}(1)=-1
\)
Q4: If \(y=\tan ^{-1} \frac{1}{1+x+x^2}+\tan ^{-1} \frac{1}{x^2+3 x+3} +\tan ^{-1} \frac{1}{x^2+5 x+7}+\cdots+\) upto \(n\) terms, then \(y^{\prime}(0)\) is
a. \(-\frac{1}{1+n^2}\)
b. \(-\frac{n^2}{1+n^2}\)
c. \(\frac{n}{1+n^2}\)
d. None of these
Solution: (b) \(y=\tan ^{-1} \frac{1}{1+x+x^2}+\tan ^{-1} \frac{1}{x^2+3 x+3}+\cdots+n \text { terms }\)
\(
=\tan ^{-1} \frac{(x+1)-x}{1+x(1+x)}+\tan ^{-1} \frac{(x+2)-(x+1)}{1+(x+1)(x+2)}+\cdots+n \text { terms }
\)
\(
\begin{gathered}
=\tan ^{-1}(x+1)-\tan ^{-1} x+\tan ^{-1}(x+2)-\tan ^{-1}(x+1) \\
+\ldots+\tan ^{-1}(x+n)-\tan ^{-1}(x+(n-1)) \\
=\tan ^{-1}(x+n)-\tan ^{-1} x
\end{gathered}
\)
\(
\begin{aligned}
& y^{\prime}(x)=\frac{1}{1+(x+n)^2}-\frac{1}{1+x^2} \\
& \Rightarrow y^{\prime}(0)=\frac{1}{1+n^2}-1=\frac{-n^2}{1+n^2}
\end{aligned}
\)
Q5: If \(\sqrt{\left(1-x^6\right)}+\sqrt{\left(1-y^6\right)}=a\left(x^3-y^3\right)\), and \(\frac{d y}{d x}=f(x, y) \sqrt{\left(\frac{1-y^6}{1-x^6}\right)}\), then .
a. \(f(x, y)=y / x\)
b. \(f(x, y)=y^2 / x^2\)
c. \(f(x, y)=2 y^2 / x^2\)
d. \(f(x, y)=x^2 / y^2\)
Solution: (d) Let \(x^3=\cos p\) and \(y^3=\cos q\)
\(
\begin{aligned}
& \text { Given } \sqrt{\left(1-x^6\right)}+\sqrt{\left(1-y^6\right)}=a\left(x^3-y^3\right) \\
& \Rightarrow \sqrt{\left(1-\cos ^2 p\right)}+\sqrt{\left(1-\cos ^2 q\right)}=a(\cos p-\cos q) \\
& \Rightarrow \sin p+\sin q=a(\cos p-\cos q) \\
& \Rightarrow 2 \sin \left(\frac{p+q}{2}\right) \cos \left(\frac{p-q}{2}\right) =-2 a \sin \left(\frac{p-q}{2}\right) \sin \left(\frac{p+q}{2}\right)
\end{aligned}
\)
\(
\begin{aligned}
& \Rightarrow \tan \left(\frac{p-q}{2}\right)=-\frac{1}{a} \\
& \Rightarrow p-q=\tan ^{-1}\left(-\frac{1}{a}\right) \\
& \Rightarrow \cos ^{-1} x^3-\cos ^{-1} y^3=\tan ^{-1}\left(-\frac{1}{a}\right)
\end{aligned}
\)
Differentiate w.r.t. \(x\), we have \(-\frac{3 x^2}{\sqrt{1-x^6}}+\frac{3 y^2}{\sqrt{1-y^6}} \frac{d y}{d x}=0\)
\(
\Rightarrow \frac{d y}{d x}=\frac{x^2}{y^2} \sqrt{\frac{1-y^6}{1-x^6}} .
\)
Hence, \(f(x, y)=x^2 / y^2\)
Q6: If \(f(x)=\cos x \cdot \cos 2 x \cdot \cos 4 x \cdot \cos 8 x \cdot \cos 16 x\), then \(f^{\prime}\left(\frac{\pi}{4}\right)\) is
a. \(\sqrt{2}\)
b. \(\frac{1}{\sqrt{2}}\)
c. 1
d. None of these
Solution: (a)
\(
\begin{aligned}
f(x) & =\frac{2 \sin x \cos x \cos 2 x \cos 4 x \cos 8 x \cos 16 x}{2 \sin x} \\
& =\frac{\sin 32 x}{2^5 \sin x} \\
\Rightarrow f^{\prime}(x) & =\frac{1}{32} \times \frac{32 \cos 32 x \sin x-\cos x \sin 32 x}{\sin ^2 x} \\
\Rightarrow f^{\prime}\left(\frac{\pi}{4}\right) & =\frac{32 \times \frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}} \times 0}{32\left(\frac{1}{\sqrt{2}}\right)^2}=\sqrt{2}
\end{aligned}
\)
Q7: If \(f(x)=|x|^{|\sin x|}\), then \(f^{\prime}\left(-\frac{\pi}{4}\right)\) equals
a. \(\left(\frac{\pi}{4}\right)^{1 / \sqrt{2}}\left(\frac{\sqrt{2}}{2} \ln \frac{4}{\pi}-\frac{2 \sqrt{2}}{\pi}\right)\)
b. \(\left(\frac{\pi}{4}\right)^{1 / \sqrt{2}}\left(\frac{\sqrt{2}}{2} \ln \frac{4}{\pi}+\frac{2 \sqrt{2}}{\pi}\right)\)
c. \(\left(\frac{\pi}{4}\right)^{1 / \sqrt{2}}\left(\frac{\sqrt{2}}{2} \ln \frac{\pi}{4}-\frac{2 \sqrt{2}}{\pi}\right)\)
d. \(\left(\frac{\pi}{4}\right)^{1 / \sqrt{2}}\left(\frac{\sqrt{2}}{2} \ln \frac{\pi}{4}+\frac{2 \sqrt{2}}{\pi}\right)\)
Solution: (a) In the neighbourhood of \(-\pi / 4\), we have
\(
\begin{aligned}
& f(x)=(-x)^{-\sin x}=e^{-\sin x \log (-x)} \\
& \Rightarrow f^{\prime}(x)=e^{-\sin x \log (-x)}\left(-\cos x \log (-x)-\frac{\sin x}{x}\right) \\
& \Rightarrow f^{\prime}(x)=(-x)^{-\sin x}\left(-\cos x \log (-x)-\frac{\sin x}{x}\right) \\
& \Rightarrow f^{\prime}(-\pi / 4)=\left(\frac{\pi}{4}\right)^{1 / \sqrt{2}}\left(\frac{-1}{\sqrt{2}} \log \frac{\pi}{4}+\frac{4}{\pi} \times \frac{-1}{\sqrt{2}}\right) \\
& =\left(\frac{\pi}{4}\right)^{1 / \sqrt{2}}\left(\frac{\sqrt{2}}{2} \log \frac{4}{\pi}-\frac{2 \sqrt{2}}{\pi}\right)
\end{aligned}
\)
Q8: If \(y=\sqrt{(a-x)(x-b)}-(a-b) \tan ^{-1} \sqrt{\frac{a-x}{x-b}}\), then \(\frac{d y}{d x}\) is equal to
a. 1
b. \(\sqrt{\frac{a-x}{x-b}}\)
c. \(\sqrt{(a-x)(x-b)}\)
d \(\frac{1}{\sqrt{(a-x)(b-x)}}\)
Solution: (b) Let \(x=a \cos ^2 \theta+b \sin ^2 \theta\)
\(\therefore a-x=a-a \cos ^2 \theta-b \sin ^2 \theta=(a-b) \sin ^2 \theta\), and
\(
\begin{aligned}
& x-b=a \cos ^2 \theta+b \sin ^2 \theta-b=(a-b) \cos ^2 \theta \\
& \therefore \quad y=(a-b) \sin \theta \cos \theta-(a-b) \tan ^{-1} \tan \theta \\
& \quad=\frac{a-b}{2} \sin 2 \theta-(a-b) \theta \\
& \therefore \frac{d y}{d x}=\frac{d y / d \theta}{d x / d \theta}=\frac{(a-b) \cos 2 \theta-(a-b)}{(b-a) \sin 2 \theta}=\frac{1-\cos 2 \theta}{\sin 2 \theta} \\
& \quad=\tan \theta=\sqrt{\frac{a-x}{x-b}}
\end{aligned}
\)
Q9: If \(x=a \cos \theta, y=b \sin \theta\), then \(\frac{d^3 y}{d x^3}\) is
a. \(-\frac{3 b}{a^3} \operatorname{cosec}^4 \theta \cot ^4 \theta\)
b. \(\frac{3 b}{a^3} \operatorname{cosec}^4 \theta \cot \theta\)
c. \(-\frac{3 b}{a^3} \operatorname{cosec}^4 \theta \cot \theta\)
d. None of these
Solution: (c) We have \(y=b \sin \theta, x=a \cos \theta\).
Therefore, \(\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}=-\frac{b}{a} \cot \theta\)
\(
\begin{aligned}
\frac{d^2 y}{d x^2} & =\frac{d}{d x}\left(\frac{d y}{d x}\right)=\frac{d}{d \theta}\left(-\frac{b}{a} \cot \theta\right) \cdot \frac{d \theta}{d x} \\
& =\frac{b}{a} \operatorname{cosec}^2 \theta \frac{d \theta}{d x}=-\frac{b}{a^2} \operatorname{cosec}^3 \theta \\
\frac{d^3 y}{d x^3} & =\frac{d}{d x}\left(\frac{d^2 y}{d x^2}\right)=\frac{d}{d \theta}\left(-\frac{b}{a^2} \operatorname{cosec}^3 \theta\right) \frac{d \theta}{d x} \\
& =-\frac{b}{a^2} 3 \operatorname{cosec}^2 \theta(-\operatorname{cosec} \theta \cot \theta) \frac{d \theta}{d x} \\
& =\frac{3 b}{a^2} \operatorname{cosec}{ }^3 \theta \cot \theta \times \frac{-1}{a \sin \theta}=-\frac{3 b}{a^3} \operatorname{cosec}^4 \theta \cot \theta
\end{aligned}
\)
Q10: If \(f(x)=\left|\begin{array}{lll}(1+x)^{a_1 b_1} & (1+x)^{a_1 b_2} & (1+x)^{a_1 b_3} \\ (1+x)^{a_2 b_1} & (1+x)^{a_2 b_2} & (1+x)^{a_2 b_3} \\ (1+x)^{a_3 b_1} & (1+x)^{a_3 b_2} & (1+x)^{a_3 b_3}\end{array}\right|\), then the coefficient of \(x\) in the expansion of \(f(x)\) is
a. 1
b 0
c. -1
d. 2
Solution: (b) We have \(f(x)=\left|\begin{array}{lll}(1+x)^{a_1 b_1} & (1+x)^{a_1 b_2} & (1+x)^{a_1 b_3} \\ (1+x)^{a_2 b_1} & (1+x)^{a_2 b_2} & (1+x)^{a_2 b_3} \\ (1+x)^{a_3 b_1} & (1+x)^{a_3 b_2} & (1+x)^{a_3 b_3}\end{array}\right|\)
\(
=a_0+a_1 x+a_2 x^2+\cdots
\)
\(
\begin{aligned}
\Rightarrow a_1=f^{\prime}(0)=\left|\begin{array}{ccc}
a_1 b_1 & a_1 b_2 & a_1 b_3 \\
1 & 1 & 1 \\
1 & 1 & 1
\end{array}\right| & +\left|\begin{array}{ccc}
1 & 1 & 1 \\
a_2 b_1 & a_2 b_2 & a_2 b_3 \\
1 & 1 & 1
\end{array}\right| \\
& +\left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & 1 & 1 \\
a_3 b_1 & a_3 b_2 & a_3 b_3
\end{array}\right|=0
\end{aligned}
\)
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