Miscellaneous Exercise (Revised) - Chapter 13 - Limits & Derivatives - Ncert Solutions class 11 - Maths
Updated On 26-08-2025 By Lithanya
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Chapter 12: Limits & Derivatives - NCERT Solutions for Class 11 Maths
Miscellaneous Exercise Question 1.
Find the derivative of following functions from first principle:
(i) $-x$
(ii) $(-x)^{-1}$
(iii) $\sin (x+1)$
(iv) $\cos \left(x-\frac{\pi}{8}\right)$
Answer.
(i) Given: $f(x)=-x$ then $f(x+h)=-(x+h)$
$
\because f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}
$
we get $f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{-(x+h)-(-x)}{h}$
$
=\lim _{h \rightarrow 0} \frac{-x-h+x}{h}=\lim _{h \rightarrow 0} \frac{-h}{h}=-1
$
(ii) Given: $f(x)=(-x)^{-1}=\frac{-1}{x}$
then $f(x+h)=\frac{-1}{x+h}$
We have $f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$
$
\begin{aligned}
& \therefore f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{\frac{-1}{x+h}-\frac{-1}{x}}{h} \\
& =\lim _{h \rightarrow 0} \frac{-x+x+h}{h x(x+h)}=\lim _{h \rightarrow 0} \frac{h}{h x(x+h)}=\frac{1}{x^2}
\end{aligned}
$
(iii)Given: $f(x)=\sin (x+1)$
then $f(x+h)=\sin (x+h+1)$
$
\because f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}
$
we get $f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{\sin (x+h+1)-\sin (x+1)}{h}$ $\left[\sin C-\sin D=2 \cos \left(\frac{C+D}{2}\right) \sin \left(\frac{C-D}{2}\right)\right]$
$=\lim _{h \rightarrow 0} \frac{2 \cos \left(\frac{2 x+h+2}{2}\right) \sin \left(\frac{h}{2}\right)}{h}$
$
=\lim _{h \rightarrow 0} \frac{\cos \left(x+1+\frac{h}{2}\right) \sin \left(\frac{h}{2}\right)}{\frac{h}{2}}
$
$
\begin{aligned}
& =\lim _{h \rightarrow 0} \cos \left(x+1+\frac{h}{2}\right) \lim _{\frac{h}{2} \rightarrow 0}\left(\frac{\sin \frac{h}{2}}{\frac{h}{2}}\right) \quad\left[\lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta}=1\right]\left[h \rightarrow 0 \Rightarrow \frac{h}{2} \rightarrow 0\right] \\
& =\cos (x+1)
\end{aligned}
$
(iv)Given: $f(x)=\cos \left(x-\frac{\pi}{8}\right)$
$\begin{aligned}
& \text { then } f(x+h)=\cos \left(x+h-\frac{\pi}{8}\right) \\
& \because f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\
& \therefore \quad f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{\cos \left(x+h-\frac{\pi}{8}\right)-\cos \left(x-\frac{\pi}{8}\right)}{h} \\
& {\left[\cos C-\cos D=-2 \sin \left(\frac{C+D}{2}\right) \sin \left(\frac{C-D}{2}\right)\right]} \\
& =\lim _{h \rightarrow 0} \frac{-2 \sin \left(x-\frac{\pi}{8}+\frac{h}{2}\right) \sin \left(\frac{h}{2}\right)}{h} \\
& =\lim _{h \rightarrow 0} \frac{-\sin \left(x-\frac{\pi}{8}+\frac{h}{2}\right) \sin \left(\frac{h}{2}\right)}{\frac{h}{2}} \\
& =\lim _{h \rightarrow 0}-\sin \left(x-\frac{\pi}{8}+\frac{h}{2}\right) \lim _{\frac{h}{2} \rightarrow 0}\left(\frac{\sin \frac{h}{2}}{\frac{h}{2}}\right) \quad\left[\lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta}=1\right]\left[h \rightarrow 0 \Rightarrow \frac{h}{2} \rightarrow 0\right] \\
&
\end{aligned}$
$
=-\sin \left(x-\frac{\pi}{8}\right)
$
Find the derivative of the following functions (it is to be understood that $a, b, c, d, p, q, r$ and $s$ are fixed non-zero constants and $m$ and $n$ are integers).
Miscellaneous Exercise Question 2.
$(x+a)$
Answer.
Given: $f(x)=(x+a)$
$
\therefore f^{\prime}(x)=\frac{d}{d x}(x+a)=\frac{d}{d x}(x)+\frac{d}{d x}(a)=1+0=1
$
Miscellaneous Exercise Question 3.
$(p x+q)\left(\frac{r}{x}+s\right)$
Answer.
Given: $f(x)=(p x+q)\left(\frac{r}{x}+s\right)$
$
\begin{aligned}
& \therefore f^{\prime}(x)=\frac{d}{d x}\left[(p x+q)\left(\frac{r}{x}+s\right)\right] \quad \text { [product rule] } \\
& \Rightarrow f^{\prime}(x)=(p x+q) \frac{d}{d x}\left(\frac{r}{x}+s\right)+\left(\frac{r}{x}+s\right) \frac{d}{d x}(p x+q) \quad \frac{d}{d x}\left(x^{-1}\right)=-1 x^{-2} \\
& =(p x+q)\left(\frac{-r}{x^2}\right)+\left(\frac{r}{x}+s\right)(p) \\
& =\frac{-p r}{x}+\frac{-q^r}{x^2}+\frac{p r}{x}+p s=\frac{-q r}{x^2}+p s
\end{aligned}
$
Miscellaneous Exercise Question 4.
$(a x+b)(c x+d)^2$
Answer.
Given: $f(x)=(a x+b)(c x+d)^2$
$
\begin{aligned}
& \therefore f^{\prime}(x)=\frac{d}{d x}\left[(a x+b)\left(c x+d^2\right] \quad\right. \text { [product rule] } \\
& \Rightarrow f^{\prime}(x)=(a x+b) \frac{d}{d x}(c x+d)^2+(c x+d)^2 \frac{d}{d x}(a x+b) \\
& =(a x+b) \frac{d}{d x}\left(c^2 x^2+2 c d x+d^2\right)+(c x+d)^2 \frac{d}{d x}(a x+b) \\
& =(a x+b)\left[c^2 \frac{d}{d x}\left(x^2\right)+2 c d \frac{d}{d x}(x)+\frac{d}{d x}\left(d^2\right)\right]+(c x+d)^2\left[a \frac{d}{d x}(x)+\frac{d}{d x}(b)\right] \\
& =(a x+b)\left(2 c^2 x+2 c d+0\right)+(c x+d)^2(a \cdot 1+0) \quad\left[\frac{d}{d x}\left(x^n\right)=n x^{n-1}\right] \\
& =2 c(a x+b)(c x+d)+a(c x+d)^2
\end{aligned}
$
Miscellaneous Exercise Question 5.
$\frac{a x+b}{c x+d}$
Answer.
Given: $f(x)=\frac{a x+b}{c x+d}$
$\therefore f^{\prime}(x)=\frac{d}{d x}\left(\frac{a x+b}{c x+d}\right)$
$\Rightarrow f^{\prime}(x)=\frac{(c x+d) \frac{d}{d x}(a x+b)-(a x+b) \frac{d}{d x}(c x+d)}{(c x+d)^2}$ [Quotient rule]
$\Rightarrow f^{\prime}(x)=\frac{(c x+d)(a)-(a x+b)(c)}{(c x+d)^2}$
$=\frac{a c x+a d-a c x-b c}{(c x+d)^2}=\frac{a d-b c}{(x+d)^2}$
Miscellaneous Exercise Question 6.
$\frac{1+\frac{1}{x}}{1-\frac{1}{x}}$
Answer.
Given: $f(x)=\frac{1+\frac{1}{x}}{1-\frac{1}{x}}=\frac{x+1}{x-1}$
$
\begin{aligned}
& \therefore f^{\prime}(x)=\frac{d}{d x}\left(\frac{x+1}{x-1}\right) \\
& \Rightarrow f^{\prime}(x)=\frac{(x-1) \frac{d}{d x}(x+1)-(x+1) \frac{d}{d x}(x-1)}{(x-1)^2}
\end{aligned}
$
[Quotient rule]
$
\begin{aligned}
& =\frac{(x-1) \times 1-(x+1) \times 1}{(x-1)^2} \\
& =\frac{x-1-x-1}{(x-1)^2}=\frac{-2}{(x-1)^2}, x \neq 0,1
\end{aligned}
$
Miscellaneous Exercise Question 7.
$\frac{1}{a x^2+b x+c}$
Answer.
Given: $f(x)=\frac{1}{a x^2+b x+c}$
$
\begin{aligned}
& \therefore f^{\prime}(x)=\frac{d}{d x}\left(\frac{1}{a x^2+b x+c}\right) \\
& \Rightarrow f^{\prime}(x)=\frac{\left(a x^2+b x+c\right) \frac{d}{d x}(1)-1 \cdot \frac{d}{d x}\left(a x^2+b x+c\right)}{\left(a x^2+b x+c\right)^2} \\
& \Rightarrow f^{\prime}(x)=\frac{\left(a x^2+b x+c\right)(0)-1(2 a x+b)}{\left(a x^2+b x+c\right)^2}
\end{aligned}
$
[Quotient rule]
$
=\frac{-(2 a x+b)}{\left(a x^2+b x+c\right)^2}
$
Miscellaneous Exercise Question 8.
$\frac{a x+b}{p x^2+q x+r}$
Answer.
Given: $f(x)=\frac{a x+b}{p x^2+q x+\mu}$
$
\therefore f^{\prime}(x)=\frac{d}{d x}\left(\frac{a x+b}{p x^2+q x+r}\right)
$
$
\begin{aligned}
& \Rightarrow f^{\prime}(x)=\frac{\left(p x^2+q x+r\right) \frac{d}{d x}(a x+b)-(a x+b) \frac{d}{d x}\left(p x^2+q x+r\right)}{\left(p x^2+q x+r\right)^2} \\
& \text { rule] } \\
& \Rightarrow f^{\prime}(x)=\frac{\left(p x^2+q x+r\right)(a)-(a x+b)(2 p x+q)}{\left(p x^2+q x+r\right)^2} \\
& \Rightarrow f^{\prime}(x)=\frac{a p x^2+a q x+a r-2 a p x^2-a q x-2 b p x-b q}{\left(p x^2+q x+r\right)^2} \\
& =\frac{-a p x^2-2 b p x+a r-b q}{\left(p x^2+q x+r\right)^2}
\end{aligned}
$
[Quotient rule]
Miscellaneous Exercise Question 9.
$\frac{p x^2+q x+r}{a x+b}$
Answer.
Given: $f(x)=\frac{p x^2+q x+\mu}{a x+b}$
$\therefore f^{\prime}(x)=\frac{d}{d x}\left(\frac{p x^2+q x+r}{a x+b}\right)$
$\Rightarrow f^{\prime}(x)=\frac{(a x+b) \frac{d}{d x}\left(p x^2+q x+r\right)-\left(p x^2+q x+r\right) \frac{d}{d x}(a x+b)}{(a x+b)^2}$
[Quotient rule]
$\Rightarrow f^{\prime}(x)=\frac{(a x+b)(2 p x+q)-\left(p x^2+q x+r\right)(a)}{(a x+b)^2}$
$\Rightarrow f^{\prime}(x)=\frac{2 a p x^2+a q x+2 b p x+b q-a p x^2-a q x-a r}{(a x+b)^2}$
$
\Rightarrow f^{\prime}(x)=\frac{a p x^2+2 b p x+b q-a r}{(a x+b)^2}
$
Miscellaneous Exercise Question 10.
$\frac{a}{x^4}-\frac{b}{x^2}+\cos x$
Answer.
Given: $f(x)=\frac{a}{x^4}-\frac{b}{x^2}+\cos x$
$
\begin{aligned}
& =a x^{-4}-b x^{-2}+\cos x \\
& \therefore f^{\prime}(x)=\frac{d}{d x}\left(a x^{-4}-b x^{-2}+\cos x\right) \\
& {\left[\frac{d}{d x}(\cos x)=-\sin x\right]} \\
& \Rightarrow f^{\prime}(x)=a \frac{d}{d x}\left(x^{-4}\right)-b \frac{d}{d x}\left(x^{-2}\right)+\frac{d}{d x} \cos x \\
& \Rightarrow f^{\prime}(x)=-4 a x^{-5}+2 b x^{-3}-\sin x
\end{aligned}
$
Miscellaneous Exercise Question 11.
$4 \sqrt{x}-2$
Answer.
Given: $f(x)=4 \sqrt{x}-2$
$
\begin{aligned}
& \therefore f^{\prime}(x)=\frac{d}{d x}(4 \sqrt{x}-2) \\
& \Rightarrow f^{\prime}(x)=4 \frac{d}{d x} \sqrt{x}-\frac{d}{d x}(2) \quad\left[\frac{d}{d x}(\sqrt{x})=\frac{1}{2 \sqrt{x}}\right] \\
& =4 \times \frac{1}{2 \sqrt{x}}-0=\frac{2}{\sqrt{x}}
\end{aligned}
$
Miscellaneous Exercise Question 12.
$(a x+b)^n$
Answer.
Given: $f(x)=(a x+b)^n$
$
\begin{aligned}
& \therefore f^{\prime}(x)=\frac{d}{d x}\left[(a x+b)^n\right] \\
& \Rightarrow f^{\prime}(x)=n(a x+b)^{n-1} \times \frac{d}{d x}(a x+b) \\
& =n(a x+b)^{n-1} \times(a)=n a(a x+b)^{n-1}
\end{aligned}
$
Miscellaneous Exercise Question 13.
$(a x+b)^n(c x+d)^m$
Answer.
Given: $(a x+b)^n(c x+d)^m$
$
\therefore f^{\prime}(x)=\frac{d}{d x}\left[(a x+b)^n\left(c x+d^2\right)^m\right]
$
[product rule]
$
\begin{aligned}
& \Rightarrow f^{\prime}(x)=(a x+b)^n \frac{d}{d x}(c x+d)^m+(c x+d)^m \frac{d}{d x}(a x+b)^n \\
& {\left[\frac{d}{d x}\left(x^n\right)=n x^{n-1}\right]} \\
& \Rightarrow f^{\prime}(x)=(a x+b)^n \cdot m(c x+d)^{m-1} \frac{d}{d x}(c x+d)+(c x+d)^m n(a x+b)^{n-1} \frac{d}{d x}(a x+b) \\
& \Rightarrow f^{\prime}(x)=(a x+b)^n \cdot m(c x+d)^{m-1}(c)+(c x+d)^m n(a x+b)^{n-1}(a) \\
& \Rightarrow f^{\prime}(x)=c m(a x+b)^n(c x+d)^{m-1}+a n(c x+d)^m(a x+b)^{n-1} \\
& \Rightarrow f^{\prime}(x)=(a x+b)^{n-1}(c x+d)^{m-1}[c m(a x+b)+a n(c x+d)]
\end{aligned}
$
Miscellaneous Exercise Question 14.
$\sin (x+a)$
Answer.
Given: $f(x)=\sin (x+a)$
$
\begin{aligned}
& \therefore f^{\prime}(x)=\frac{d}{d x}[\sin (x+a)] \\
& \Rightarrow f^{\prime}(x)=\cos (x+a)-\frac{d}{d x}(x+a)=\cos (x+a)
\end{aligned}
$
Miscellaneous Exercise Question 15.
$\operatorname{cosec} x \cot x$
Answer.
Given: $f(x)=\operatorname{cosec} x \cot x$
$
\begin{aligned}
& \therefore f^{\prime}(x)=\frac{d}{d x}(\operatorname{cosec} x \cot x) \\
& \Rightarrow f^{\prime}(x)=\operatorname{cosec} x \frac{d}{d x} \cot x+\cot x \frac{d}{d x} \operatorname{cosec} x \\
& \Rightarrow f^{\prime}(x)=\operatorname{cosec} x\left(-\operatorname{cosec}^2 x\right)+\cot x(-\operatorname{cosec} x \cot x) \\
& =-\operatorname{cosec} 3-\operatorname{cosec} x \cdot \cot ^2 x \\
& =-\operatorname{cosec} x\left(\operatorname{cosec} x+\cot ^2 x\right)
\end{aligned}
$
Miscellaneous Exercise Question 16.
$\frac{\cos x}{1+\sin x}$
Answer.
Given: $f(x)=\frac{\cos x}{1+\sin x}$
$\begin{aligned}
& \therefore f^{\prime}(x)=\frac{d}{d x}\left(\frac{\cos x}{1+\sin x}\right) \\
& \Rightarrow f^{\prime}(x)=\frac{(1+\sin x) \frac{d}{d x} \cos x-\cos x \frac{d}{d x}(1+\sin x)}{(1+\sin x)^2} \quad\left[\frac{d}{d x}(\cos x)=-\sin x\right]
\end{aligned}$
$\begin{aligned}
& =\frac{(1+\sin x)(-\sin x)-\cos x(\cos x)}{(1+\sin x)^2} \\
& {\left[\frac{d}{d x}(\sin x)=\cos x\right]} \\
& \Rightarrow f^{\prime}(x)=\frac{-\sin x-\sin ^2 x-\cos ^2 x}{(1+\sin x)^2} \\
& =\frac{-\sin x-\left(\sin ^2 x+\cos ^2 x\right)}{(1+\sin x)^2}=\frac{-\sin x-1}{(1+\sin x)^2} \\
& {\left[\sin ^2 x+\cos ^2 x=1\right]} \\
& \Rightarrow f^{\prime}(x)=\frac{-(1+\sin x)}{(1+\sin x)^2}=\frac{-1}{1+\sin x} \\
&
\end{aligned}$
Miscellaneous Exercise Question 17.
$\frac{\sin x+\cos x}{\sin x-\cos x}$
Answer.
Given: $f(x)=\frac{\sin x+\cos x}{\sin x-\cos x}$
$\therefore f^{\prime}(x)=\frac{d}{d x}\left(\frac{\sin x+\cos x}{\sin x-\cos x}\right)$
$\Rightarrow f^{\prime}(x)=\frac{(\sin x-\cos x) \frac{d}{d x}(\sin x+\cos x)-(\sin x+\cos x) \frac{d}{d x}(\sin x-\cos x)}{(\sin x-\cos x)^2}$
$\Rightarrow \quad f^{\prime}(x)=\frac{(\sin x-\cos x)(\cos x-\sin x)-(\sin x+\cos x)(\cos x+\sin x)}{(\sin x-\cos x)^2}$
$\left[\frac{d}{d x}(\cos x)=-\sin x\right]$
$\Rightarrow f^{\prime}(x)=\frac{-\sin ^2 x-\cos ^2 x+2 \sin x \cos x-\sin ^2 x-\cos ^2 x-2 \sin x \cos x}{(\sin x-\cos x)^2}$
$\left[\frac{d}{d x}(\sin x)=\cos x\right]$
$\begin{array}{ll}
\Rightarrow f^{\prime}(x)=\frac{-2\left(\sin ^2 x+\cos ^2 x\right)}{(\sin x-\cos x)^2} & {\left[\sin ^2 x+\cos ^2 x=1\right]} \\
=\frac{-2}{(\sin x-\cos x)^2} &
\end{array}$
Miscellaneous Exercise Question 18.
$\frac{\sec x-1}{\sec x+1}$
Answer.
Given: $f(x)=\frac{\sec x-1}{\sec x+1}$
$
\begin{aligned}
& \therefore f^{\prime}(x)=\frac{d}{d x}\left(\frac{\sec x-1}{\sec x+1}\right) \\
& \Rightarrow f^{\prime}(x)=\frac{(\sec x+1) \frac{d}{d x}(\sec x-1)-(\sec x-1) \frac{d}{d x}(\sec x+1)}{(\sec x+1)^2} \\
& \Rightarrow f^{\prime}(x)=\frac{(\sec x+1)(\sec x \tan x)-(\sec x-1)(\sec x \tan x)}{(\sec x+1)^2} \\
& \Rightarrow f^{\prime}(x)=\frac{\sec x \tan x+\sec x \tan x-\sec x \tan x+\sec x \tan x}{(\sec x+1)^2} \\
& \Rightarrow f^{\prime}(x)=\frac{2 \sec x \tan x}{(\sec x+1)^2}
\end{aligned}
$
Miscellaneous Exercise Question 19.
$\sin ^n x$
Answer.
Given: $f(x)=\sin ^n x$
$
\begin{aligned}
& \therefore f^{\prime}(x)=\frac{d}{d x}\left(\sin ^n x\right) \\
& \Rightarrow f^{\prime}(x)=n \sin ^{n-1} x \frac{d}{d x}(\sin x) \quad\left[\frac{d}{d x}(\sin x)=\cos x\right] \\
& =n \sin ^{n-1} x \cos x
\end{aligned}
$
Miscellaneous Exercise Question 20.
$\frac{a+b \sin x}{c+d \cos x}$
Answer.
Given: $f(x)=\frac{a+b \sin x}{c+d \cos x}$
$
\begin{aligned}
& \therefore f^{\prime}(x)=\frac{d}{d x}\left(\frac{a+b \sin x}{c+d \cos x}\right) \\
& \Rightarrow \quad f^{\prime}(x)=\frac{(c+d \cos x) \frac{d}{d x}(a+b \sin x)-(a+b \sin x) \frac{d}{d x}(c+d \cos x)}{(c+d \cos x)^2} \\
& {\left[\frac{d}{d x}(\sin x)=\cos x\right]}
\end{aligned}
$
$
\begin{aligned}
& =\frac{b c \cos x+a d \sin x+b d\left(\cos ^2 x+\sin ^2 x\right)}{(c+d \cos x)^2} \quad\left[\sin ^2 x+\cos ^2 x=1\right] \\
& =\frac{b c \cos x+a d \sin x+b d}{(c+d \cos x)^2}
\end{aligned}
$
Miscellaneous Exercise Question 21.
$\frac{\sin (x+a)}{\cos x}$
Answer.
Given: $f(x)=\frac{\sin (x+a)}{\cos x}$
$
\begin{aligned}
& \therefore f^{\prime}(x)=\frac{d}{d x}\left[\frac{\sin (x+a)}{\cos x}\right] \\
& \Rightarrow f^{\prime}(x)=\frac{\cos x \frac{d}{d x}[\sin (x+a)]-\sin (x+a) \frac{d}{d x} \cos x}{\cos ^2 x} \\
& =\frac{\cos x \cdot \cos (x+a) \cdot \frac{d}{d x}(x+a)-\sin (x+a) \cdot \frac{d}{d x}(\cos x)}{\cos ^2 x} \\
& \Rightarrow f^{\prime}(x)=\frac{\cos x \cos (x+a)-\sin (x+a)(-\sin x)}{\cos ^2 x} \\
& \Rightarrow \quad f^{\prime}(x)=\frac{\cos x \cos (x+a)+\sin x \sin (x+a)}{\cos ^2 x} \\
& {[\cos (A-B)=\cos A \cdot \cos B+\sin A \cdot \sin B]} \\
& =\frac{\cos (x+a-x)}{\cos ^2 x}=\frac{\cos a}{\cos ^2 x} \\
&
\end{aligned}
$
Miscellaneous Exercise Question 22.
$x^4(5 \sin x-3 \cos x)$
Answer.
Given: $f(x)=x^4(5 \sin x-3 \cos x)$
$
\therefore f^{\prime}(x)=\frac{d}{d x}\left[x^4(5 \sin x-3 \cos x)\right]
$
[product rule]
$\begin{aligned}
& \Rightarrow f^{\prime}(x)=x^4 \frac{d}{d x}(5 \sin x-3 \cos x)+(5 \sin x-3 \cos x) \frac{d}{d x}\left(x^4\right) \quad\left[\frac{d}{d x}(\sin x)=\cos x\right] \\
& \Rightarrow f^{\prime}(x)=x^4(5 \cos x+3 \sin x)+(5 \sin x-3 \cos x)\left(4 x^3\right) \quad\left[\frac{d}{d x}(\cos x)=-\sin x\right]
\end{aligned}$
$
\Rightarrow f^{\prime}(x)=x^3(5 x \cos x+3 x \sin x+20 \sin x-12 \cos x) \quad\left[\frac{d}{d x}\left(x^n\right)=n x^{n-1}\right]
$
Miscellaneous Exercise Question 23.
$\left(x^2+1\right) \cos x$
Answer.
Given: $f(x)=\left(x^2+1\right) \cos x$
$
\begin{array}{ll}
\therefore f^{\prime}(x)=\frac{d}{d x}\left[\left(x^2+1\right) \cos x\right] & \text { [product rule] } \\
\Rightarrow f^{\prime}(x)=\left(x^2+1\right) \frac{d}{d x} \cos x+\cos x \frac{d}{d x}\left(x^2+1\right) & {\left[\frac{d}{d x}(\cos x)=-\sin x\right]}
\end{array}
$
$
\begin{aligned}
& =\left(x^2+1\right)(-\sin x)+\cos x(2 x) \quad\left[\frac{d}{d x}\left(x^n\right)=n x^{n-1}\right] \\
& \Rightarrow f^{\prime}(x)=-x^2 \sin x-\sin x+2 x \cos x
\end{aligned}
$
Miscellaneous Exercise Question 24.
$\left(a x^2+\sin x\right)(p+q \cos x)$
Answer.
Given: $f(x)=\left(a x^2+\sin x\right)(p+q \cos x)$
$
\therefore f^{\prime}(x)=\frac{d}{d x}\left[\left(a x^2+\sin x\right)(p+q \cos x)\right]
$
[product rule]
$
\begin{aligned}
& \Rightarrow \quad f^{\prime}(x)=\left(a x^2+\sin x\right) \frac{d}{d x}(p+q \cos x)+(p+q \cos x) \frac{d}{d x}\left(a x^2+\sin x\right) \\
& {\left[\frac{d}{d x}(\sin x)=\cos x\right]} \\
& \Rightarrow f^{\prime}(x)=\left(a x^2+\sin x\right)(-q \sin x)+(p+q \cos x)(2 a x+\cos x) \\
& {\left[\frac{d}{d x}(\cos x)=-\sin x\right]} \\
& \Rightarrow f^{\prime}(x)=-q \sin x\left(a x^2+\sin x\right)+(p+q \cos x)(2 a x+\cos x)
\end{aligned}
$
Miscellaneous Exercise Question 25.
$(x+\cos x)(x-\tan x)$
Answer.
Given: $f(x)=(x+\cos x)(x-\tan x)$
$
\begin{aligned}
& \therefore f^{\prime}(x)=\frac{d}{d x}[(x+\cos x)(x-\tan x)] \\
& \text { [product rule] } \\
& \Rightarrow \quad f^{\prime}(x)=(x+\cos x) \frac{d}{d x}(x-\tan x)+(x-\tan x) \frac{d}{d x}(x+\cos x) \\
& {\left[\frac{d}{d x}(\tan x)=\sec ^2 x\right]} \\
& \Rightarrow f^{\prime}(x)=(x+\cos x)\left(1-\sec ^2 x\right)+(x-\tan x)(1-\sin x) \\
& {\left[\frac{d}{d x}(\cos x)=-\sin x\right]} \\
& \Rightarrow f^{\prime}(x)=-\tan ^2 x(x+\cos x)+(x-\tan x)(1-\sin x) \quad\left[\sec ^2 \theta-\tan ^2 \theta=1\right] \\
&
\end{aligned}
$
Miscellaneous Exercise Question 26.
$\frac{4 x+5 \sin x}{3 x+7 \cos x}$
Answer.
Given: $f(x)=\frac{4 x+5 \sin x}{3 x+7 \cos x}$
$\therefore f^{\prime}(x)=\frac{d}{d x}\left(\frac{4 x+5 \sin x}{3 x+7 \cos x}\right)$
$\Rightarrow f^{\prime}(x)=\frac{(3 x+7 \cos x) \frac{d}{d x}(4 x+5 \sin x)-(4 x+5 \sin x) \frac{d}{d x}(3 x+7 \cos x)}{(3 x+7 \cos x)^2}$ [Quotient
rule]
$\Rightarrow f^{\prime}(x)=\frac{(3 x+7 \cos x)(4+5 \cos x)-(4 x+5 \sin x)(3-7 \sin x)}{(3 x+7 \cos x)^2}$
$
\begin{aligned}
& \Rightarrow \\
& f^{\prime}(x)=\frac{12 x+15 x \cos x+28 \cos x+35 \cos ^2 x-12 x+28 x \sin x+35 \sin ^2 x-15 \sin x}{(3 x+7 \cos x)^2}
\end{aligned}
$
$\begin{aligned}
& \Rightarrow \quad f^{\prime}(x)=\frac{15 x \cos x+28 \cos x+28 x \sin x-15 \sin x+35\left(\cos ^2 x+\sin ^2 x\right)}{(3 x+7 \cos x)^2} \\
& {\left[\sin ^2 x+\cos ^2 x=1\right]} \\
& \Rightarrow f^{\prime}(x)=\frac{15 x \cos x+28 \cos x+28 x \sin x-15 \sin x+35}{(3 x+7 \cos x)^2}
\end{aligned}$
Miscellaneous Exercise Question 27.
$\frac{x^2 \cos \left(\frac{\pi}{4}\right)}{\sin x}$
Answer.
Given:
$
f(x)=\frac{x^2 \cos \left(\frac{\pi}{4}\right)}{\sin x}
$
$\therefore f^{\prime}(x)=\frac{d}{d x}\left[\frac{x^2 \cos \left(\frac{\pi}{4}\right)}{\sin x}\right]$
[Quotient rule]
$
\begin{aligned}
& \Rightarrow f^{\prime}(x)=\frac{\sin x \frac{d}{d x}\left[x^2 \cos \left(\frac{\pi}{4}\right)\right]-x^2 \cos \left(\frac{\pi}{4}\right) \frac{d}{d x}(\sin x)}{\sin ^2 x} \quad\left[\frac{d}{d x}(\sin x)=\cos x\right] \\
& \Rightarrow f^{\prime}(x)=\frac{\sin x\left[2 x \cos \left(\frac{\pi}{4}\right)\right]-x^2 \cos \left(\frac{\pi}{4}\right)(\cos x)}{\sin ^2 x} \quad\left[\frac{d}{d x}\left(x^n\right)=n x^{n-1}\right] \\
& \Rightarrow f^{\prime}(x)=\frac{2 x \sin x \cos \left(\frac{\pi}{4}\right)-x^2 \cos x \cos \left(\frac{\pi}{4}\right)}{\sin ^2 x}
\end{aligned}
$
$
=f^{\prime}(x)=\frac{x \cos \left(\frac{\pi}{4}\right)[2 \sin x-x \cos x]}{\sin ^2 x}
$
Miscellaneous Exercise Question 28.
$\frac{x}{1+\tan x}$
Answer.
Given: $f(x)=\frac{x}{1+\tan x}$
$
\begin{aligned}
& \therefore f^{\prime}(x)=\frac{d}{d x}\left(\frac{x}{1+\tan x}\right) \\
& \Rightarrow f^{\prime}(x)=\frac{(1+\tan x) \frac{d}{d x}(x)-x \frac{d}{d x}(1+\tan x)}{(1+\tan x)^2} \quad\left[\frac{d}{d x}(\tan x)=\sec ^2 x\right]
\end{aligned}
$
$
\begin{aligned}
& =\frac{(1+\tan x)(1)-x\left(\sec ^2 x\right)}{(1+\tan x)^2} \\
& \Rightarrow f^{\prime}(x)=\frac{1+\tan x-x \sec ^2 x}{(1+\tan x)^2}
\end{aligned}
$
Miscellaneous Exercise Question 29.
$(x+\sec x)(x-\tan x)$
Answer.
Given: $f(x)=(x+\sec x)(x-\tan x)$
$
\therefore f^{\prime}(x)=\frac{d}{d x}[(x+\sec x)(x-\tan x)]
$
[product rule]
$\begin{aligned}
& \Rightarrow f^{\prime}(x)=(x+\sec x) \frac{d}{d x}(x-\tan x)+(x-\tan x) \frac{d}{d x}(x+\sec x) \\
& {\left[\frac{d}{d x}(\tan x)=\sec ^2 x\right]}
\end{aligned}$
$\begin{aligned}
& =(x+\sec x)\left(1-\sec ^2 x\right)+(x-\tan x)(1+\sec x \tan x) \\
& {\left[\frac{d}{d x}(\sec x)=\sec x \tan x\right]} \\
&
\end{aligned}$
Miscellaneous Exercise Question 30.
$\frac{x}{\sin ^n x}$
Answer.
Given: $f(x)=\frac{x}{\operatorname{sint}^n x}$
$\therefore f^{\prime}(x)=\frac{d}{d x}\left(\frac{x}{\sin ^7 x}\right)$
$\Rightarrow f^{\prime}(x)=\frac{\sin ^2 x \frac{d}{d x}(x)-x \frac{d}{d x}\left(\sin ^2 x\right)}{\left(\sin ^2 x\right)^2}$
$=\frac{\sin ^n x \cdot 1-x \cdot n \cdot \sin ^{n-1} x \cdot \frac{d}{d x}(\sin x)}{\left(\sin ^n x\right)^2} \quad\left[\frac{d}{d x}(\sin x)=\cos x\right]$
$\Rightarrow f^{\prime}(x)=\frac{\sin ^2 x-r x \sin ^{-1} x \cos x}{\left(\sin ^2 x\right)^2}$
$=\frac{\sin ^{n-1} x(\sin x-n x \cos x)-n x \sin ^{n-1} x \cos x}{\sin ^{2 n} x}$
$=\frac{\sin ^{n-1} x(\sin x-n x \cos x)}{\sin ^{2 n} x}=\sin ^{n-1-2 n} x(\sin x-n x \cos x)=\sin ^{-(n+1)} x(\sin x-n x \cos x)$
$=\frac{\sin x-n x \cos x}{\sin ^{n+1} x}$
