KC Sinha Solution Class 11 Chapter 26 अवकलज (Derivatives) Exercise 26.4

Exercise 26.4

निम्नलिखित फलनों का x के सापेक्ष अवकलन करें :

[Differentiate the following functions w.r.t. to x]:

Question 1

(i) $x^{3} \cos x$

Sol :

माना $y=x^{3} \cdot \cos x$

Differentiating w.r.t x

$\frac{d y}{d x}=\frac{d\left(x^{3}\right)}{d x} \cdot \cos x + x^3 \cdot \frac{d(\cos x)}{dx}$

$=3 x^{2} \cos x+x^{3}(-\sin x)$

$=3 x^{2} \cos x-x^{3} \sin x$

(ii) $x^{4} \sin x$

Sol :

माना $y=x^{4} \sin x$

$\frac{d y}{d x}=\frac{d\left(x^{4}\right)}{d x} \cdot \sin x+x^{4} \cdot \frac{d(\sin x)}{d x}$

$=4 x^{3} \sin x+x^{4} \cos x$

(iii) $\left(x^{2}+1\right)\left(x^{3}+2 x+5\right)$

Sol :

माना $y=\left(x^{2}+1\right) \cdot\left(x^{3}+2 x+5\right)$

Differentiating w.r.t x

$\frac{d y}{d x}=\frac{d\left(x^{2}+1\right)}{d x}.\left(x^{3}+2 x+5\right)+\left(x^{2}+1\right)\frac{d(x^3+2x+5)}{dx}$

$=2 x \cdot\left(x^{3}+2 x+5\right)+\left(x^{2}+1\right)\left(3 x^{2}+2\right)$

$=2 x^{4}+4 x^{2}+10 x+3 x^{4}+2x^2+3x^2+2$

$=5 x^{4}+9 x^{2}+10 x+2$

(iv) $\left(x^{2}+1\right) \sin x$

Sol :

माना $y=\left(x^{2}+1\right) \sin x$

Differentiating w.r.t x

$\frac{d y}{d x}=\frac{d\left(x^{2}+1\right)}{d x} \cdot \sin x+\left(x^{2}+1\right) \frac{d(\sin x)}{d x}$

$=2 x \sin x+\left(x^{2}+1\right) \cos x$

(v) $(p x+q)\left(\frac{r}{x}+s\right)$

Sol :

माना $y=(p x+q) \cdot\left(\frac{r}{x}+s\right)$

Differentiating w.r.t x

$\frac{d y}{d x}=\frac{d(p x-q)}{d x} \cdot\left(\frac{r}{x}+s\right)+(p x+q) \cdot \frac{d\left(\frac{r}{x}+s\right)}{d x}$

$=p \cdot\left(\frac{r}{x}+s\right)+(p x+q)\left(\frac{-r}{x^{2}}\right)$

$=\frac{p r}{2}+p s-\frac{p r}{2}-\frac{q r}{x^{2}}$

$=p s-\frac{qr}{x^{2}}$

(vi) $(a x+b)(c x+d)^{2}$

Sol :

माना $y=(a x+b)(c x+d)^{2}$

Differentiating w.r.t x

$\frac{d y}{dx}=\frac{d(a x+b)}{dx} \cdot(cx+d)^{2}+(a x+b) \cdot \frac{d(cx+d)^{2}}{d x}$

$=a(c x+d)^{2}+(a x+b) \cdot 2(cx+d).c$

$=a\left(cx+d\right)^{2}+2 c(a x+b)\left(cx+d\right)$

(vii) $\sin x \cos x$

Sol :

माना $y=\sin x \cos x$

Differentiating w.r.t x

$\frac{d y}{d x}=\frac{d(\sin x)}{d x} \cos x+\sin x \cdot \frac{d(\cos x)}{d x}$

=cos x. cos x+sin x(-sin x)

$=\cos ^{2} x-\sin ^{2} x=\cos 2 x$

(viii) $\left(x^{2}+1\right) \cos x$

Sol :

माना $y=\left(x^{2}+1\right) \cos x$

Differentiating w.r.t x

$\frac{d y}{d x}=\frac{d\left(a x^{2}+\sin x\right)}{dx} \cdot(p+q \cos x)+(ax^2+\sin x).\frac{d(p+q\cos x)}{dx}$

$=(2 a x+\cos x)(p+q \cos x)+\left(a x^{2}+\sin x \right)(-q\sin x)$

$=(2 a x+\cos x)(p+q \cos x)-q\sin x(ax^2+\sin x)$

(ix) $\left(a x^{2}+\sin x\right)(p+q \cos x)$

Sol :

(x) $x^{4}(5 \sin x-3 \cos x)$

Sol :

माना $y=x^{4}(5 \sin x-3 \cos x)$

Differentiating w.r.t x

$\frac{d y}{d x}=\frac{d\left(x^{4}\right)}{d x} \cdot(5 \sin x-3 \cos x)+x^{4} \cdot \frac{d(5 \sin x-3\cos x)}{dx}$

$=4 x^{3}(5 \sin x-3\cos x)+x^{4}(5 \cos x+3\sin x)$

$=x^{3}[4(5 \sin x-3 \cos x)+x(5 \cos x+3 \sin x)]$

$=x^{3}[20 \sin x-12 \cos x+5x \cos x+3 x \sin x]$

Question 2

(i) $x \sin x$

Sol :

माना y=x sinx 

Differentiating w.r.t x

$\frac{d y}{dx}=\frac{d(x)}{dx} \cdot \sin x+x \cdot \frac{d(\sin x)}{dx}$

=1.sin x+cosx

=sin x+xcosx

(ii) $\left(\frac{3}{2} x+7\right)\left(2 x^{2}+1\right)$

Sol :

माना $y=\left(\frac{3}{2} x+7\right)\left(2 x^{2}+1\right)$

$\frac{d y}{d x}=\frac{d\left(\frac{3}{2} x+y\right)}{d x} \cdot\left(2 x^{2}+1\right)+\left(\frac{3}{2} x+7\right) \cdot \frac{d\left(2 x^{2}+1\right)}{d}$

$=\frac{3}{2}\left(2 x^{2}+1\right)+\left(\frac{3}{2} x+7\right)(4 x)$

$=3 x^{2}+\frac{3}{2}+6 x^{2}+28 x$

$=9 x^{2}+28 x+\frac{3}{2}$

(iii) $\left(x^{2}-4 x+5\right)\left(x^{3}-2\right)$

Sol :

(iv) $x^{5} \tan x$

Sol :

(v) $\left(5+x^{2}\right) \sec x$

Sol :

(vi) $x^{3} \cot x$

Sol :

(vii) $(x+\cos x)(x-\tan x)$

Sol :

माना y=(x+cos x)(x-tanx)

Differentiating w.r.t x

$\frac{d y}{d x}=\frac{d(x+\cos x)}{dx}(x-\tan x)+(x+\cos x)\frac{d(x-\tan x)}{dx}$

=(1-\sin x)(x-\tan x)+(x+\cos x)$\left(1-\sec ^2\right)$

$=(1-\sin x)(x-\tan x)-\tan ^{2} x(x+\cos x)$

(viii) $(x+\sec x)(x-\tan x)$

Sol :

माना y=(x+sec x)(x-tan x)

Differentiating w.r.t x

$\frac{d y}{d x}=\frac{d(x+\sec x)}{dx} \cdot(x-\tan x)+(x+\sec x)\frac{d(x-tan x )}{dx}$

$=(1+\sec x \tan x)(x-\tan x)+(x+\sec x)(1-\sec ^2 x)$

Question 3

Find the derivatives of the following functions w.r.t to x

(i) $3 x^{4} \cos x+5$

Sol :

माना $y=3 x^{4} \cos x+5$

Differentiating w.r.t x

$\frac{d y}{dx}=\frac{3 \cdot d\left(x^{4} \cdot \cos x\right)}{dx}+\frac{d(5)}{dx}$

$=3\left[\frac{d\left(x^{4}\right)}{d x} \cdot \cos x+x^{4} \cdot \frac{d(\cos x)}{dx}\right]+0$

$=3\left[4 x^{3} \cos x+x^{4}(-\sin x)]\right.$

$=12 x^{3} \cos x-3 x^{4} \sin x$

(ii) $\operatorname{cosec} x \cot x$

Sol :

माना y=cosec x. cot x

Differentiating w.r.t x

$\frac{dy}{dx}=\frac{d\left(\text{cosec x} \right)}{dx} \cdot \cot x+\operatorname{cosec} x \cdot \frac{d\left(\cot x\right)}{dx}$

=-cosec x. cot x+cosec x.$\left(-\operatorname{cosc}^{2} x\right)$

$=-\operatorname{cosec} x\left(\cot ^{2} x+\operatorname{cosec}^{2} x\right)$

(iii) $x^{5} \cot x$

Sol :

माना 

Differentiating w.r.t x

(iv) $\left(x^{2}-5 x+6\right)\left(x^{3}+2\right)$

Sol :

माना

Differentiating w.r.t x

(v) $\left(x^{2}-5 x+6\right) \sec x$

Sol :

माना 

Differentiating w.r.t x

(vi) $\cos ^{2} x$

Sol :

माना $y=\cos ^{2} x$

Differentiating w.r.t x

$\frac{d y}{dx}=\frac{d(\cos x)}{dx} \cdot \cos x+\cos x \cdot \frac{dx(\cos x)}{dx}$

=-sin x.cos x+cos x(-sin x)

=-2sin x. cos x

=-sin 2x

Question 4

$\frac{d y}{d x}$ निकालें यदि (Find $\frac{d y}{d x}$ if

(i) $y=(x+1)^{3}(2 x+1)^{5}$

Sol :

$y=(x+1)^{3}(2 x+1)^{5}$

Differentiating w.r.t x

$\frac{d y}{dx}=\frac{d(x+1)^{3}}{dx} \cdot(2 x+1)^{5}+(x+1)^{3} \cdot \frac{d(2 x+1)^{5}}{dx}$

$=3(x+1)^{2} \cdot 1(2 x+1)^{5}+(x+1)^{3} \cdot 5(2 x+1)^{4} \cdot 2$

$=(x+1)^{2}(2 x+1)^{4}[3(2 x+1)+10(x+1)]$

$=(x+1)^{2}(2 x+1)^{4}[6 x+3+10x+10]$

$=(x+1)^{2}(2 x+1)^{4}(16 x+13)$

(ii) y=(1-2tan x)(5+4sin x)

Sol :

Differentiating w.r.t x

$\frac{d y}{d x}=\frac{d(1-2 \tan x)}{d x} \cdot(5+4 \sin x)+(1-2 \tan x) \frac{d\left(5+4\sin x\right)}{dx}$

$=-2 \sec ^{2}x(5+4 \sin x)+(1-2\tan x).4\cos x$

$=-2 \sec ^{2} x(5+4 \sin x)+4 \cos x(1-2 \tan x)$

Question 5

Find $\frac{d y}{d x}$ if,

$\frac{d(u \cdot v \cdot w)}{d x}=\frac{d u}{d x} \cdot v w+u \cdot \frac{d v}{d x} w+u \cdot v \cdot \frac{dw}{d n}$

(i) $y=\left(x^{2}-x+1\right)(2 x+3)\left(x^{4}+7\right)$

Sol :

$\frac{d y}{dx}=\frac{d\left(x^{2}-x-1\right)}{d x} \cdot(2 x+3)\left(x^{4}+7\right)+\left(x^{2}-x+1\right) \frac{d(2 x+3)}{dx}(x^4+7)+\left(x^{2}-x+1\right)(2 x+3) d \frac{\left(x^{4}+7\right)}{d}$

$=(2 x-1)(2 x+3)\left(x^{4}+7\right)+\left(x^{2}-x+1\right)(2)\left(x^{4}+7\right)+\left(x^{2}-x+1\right)(2 x+3) \cdot 4 x^{3}$

$=(2 x-1)(2 x+3)\left(x^{4}+7\right)+2\left(x^{2}-1+1\right)\left(x^{4}+7\right)+4 x^{3}\left(x^{2}-x+1\right)(2 x+3)$

(ii) $y=(x+1)\left(3 x^{2}+2\right)\left(5 x^{3}+3\right)$

Sol :

$y=(x+1)\left(3 x^{2}+2\right)\left(5 x^{3}+3\right)$

Differentiating w.r.t x

$\frac{d y}{d x}=\frac{d(x+1)}{d} \cdot\left(3 x^{2}+2\right) \cdot\left(5 x^{3}+3\right)+(x+1) d\left(\frac{3 x^{2}+2}{d x}\right) \cdot\left(5 x^{3}+3\right)+(x+1)\left(3 x^{2}+2\right) \frac{d\left(5 x^{3}+3\right)}{d x}$

$=1 \cdot\left(3 x^{2}+2\right)\left(5 x^{3}+3\right)+(x+1) 6 x\left(5 x^{3}+3\right)+(x+1)\left(3 x^{2}+2\right)15 x^{2}$

$=\left(3 x^{2}+2\right)\left(5 x^{3}+3\right)+6 x(2+1)\left(5 x^{3}+3\right)+15 x^{2}(x+1)\left(3 x^{2}+2\right)$

Question 6

$x^{2}$ का अवकलन गुणांक नियम से करें तथा जाँच करें कि उत्तर 2x है।

[Differentiate $x^{2}$ by product rule and verify that the answer is 2x.]

Sol :

माना $y=x^{2}$

y=x.x

Differentiating w.r.t x

$\frac{d y}{d x}=\frac{d(x)}{d x}-x+x \frac{d(x)}{d x}$

=1.x+x.1

=x+x=2x

Question 7

(3secx-4cosec x)(2sin x+5cos x) का गुणन नियम तथा अन्य विधि से अवकलन करें तथा दिखलाएँ कि दोनों विधियों से प्राप्त उत्तर समान है।

Sol :

माना y=(3sec x-4cosec x)(2sin x+5cos x)

Differentiating w.r.t x

$\frac{d y}{d x}=\frac{d(3 \sec x-4 \text{cosec x})}{d x}(2 \sin x+5 \operatorname{cosec x})+(3\sec x-4\text{cosec x})\frac{d(2\sin x+5\cos x)}{dx}$

=(3secx.tanx+4cosecx.cotx)(2sin x+5cosx)+(3secx-4cosecx)(2cosx-5sinx)

$=6 \tan ^{2} x+15\tan x+8\cot x+20\cot^2 x+6-15\tan x-8\cot x+20$

$=6 \tan ^{2} x+20 \cot ^{2} x+26$

Question 9

निम्नलिखित फलनों का गुणन नियम तथा अन्य विधि से अवकलन करें तथा जाँच करें कि दोनों उत्तर समान है।

[Differentiate the following functions by product rule and by other method and verify that the answer from both methods is the same.]

(i) $(x+2)(x+3)$

Sol :

(ii) $\left(3 x^{2}+2\right)^{2}$

Sol :

माना $y=\left(3 x^{2}+2\right)^{2}$

$y=\left(3 x^{2}+2\right)\left(3 x^{2}+2\right)$

Differentiating w.r.t x

$\frac{d y}{dx}=6 x\left(3 x^{2}+2\right)+\left(3 x^{2}+2\right) \cdot 6 x$

$=18 x^{3}+12 x+18 x^{3}+12 x$

$=36 x^{3}+24 x$

IInd method

$y=\left(3 x^{2}+2\right)^{2}$

$y=\left(3 x^{2}\right)^{2}+2 \cdot 3 x^{2} \cdot 2+2^{2}$

$y=9 x^{4}+12 x^{2}+4$

Differentiating w.r.t x

$\frac{dy}{dx}=9 \cdot 4 x^{3}+12.2 x+0$

$=36 x^{3}+24 x$

(iii) $\left(x^{2}-5 x+8\right)\left(x^{3}+7 x-9\right)$

Sol :

Question 10

We know that $\tan x \cos x=\sin x$ and $\frac{d}{d x}(\sin x)=\cos x$. Differentiate $\tan x \cos x$ by product rule and check that the answer is cos x

Sol :

$d \frac{(\tan x \cdot \cos x)}{d x}=\frac{d(\tan x)}{dx} \cdot \cos x+\tan x \cdot \frac{d(\cos x)}{dx}$

$=\sec ^{2} x \cdot \cos x+\tan x(-\sin x)$

$=\frac{1}{\cos ^{2} x} \cdot \cos x+\frac{\sin x}{\cos x}(-\sin x)$

$=\frac{1}{\cos x}-\frac{\sin^2 x}{\cos x}=\frac{1-\sin^2 x}{\cos x}$

$=\frac{\cos^{2} x}{\cos x}$

=cos x

$d\left(\frac{\tan x \cdot \cos x}{d x}\right)=\cos x$

Question 11

निम्नलिखित फलनों का x के सापेक्ष अवकलज निकालें 

[Find the derivatives of the following functions w.r.t. to x]

(i) (a) $\frac{x+1}{x-1}$

Sol :

माना $y=\frac{x+1}{x-1}$

Differentiating w.r.t x

$\frac{d y}{d x}=\frac{\frac{d(x+1)}{dx} \cdot(x-1)-(x+1) \cdot d \frac{(x-1)}{dx}}{(x-1)^{2}}$

$=\frac{1(x-1)-(x+1) \cdot 1}{(x-1)^{2}}$

$=\frac{x-1-x-1}{(x-1)^{2}}$

$=\frac{-2}{(x-1)^{2}}$

(b) $\frac{1+\frac{1}{x}}{1-\frac{1}{x}}$

Sol :

(ii) $\frac{7 x+4}{4 x-7}$

Sol :

माना $y=\frac{7 x+4}{4 x-7}$

Differentiating w.r.t x

$\frac{d y}{d x}=\frac{\frac{d(7 x+4)}{d x} \cdot(4 x-7)-(7 x+4) \cdot d \frac{(4 x-7)}{dx}}{(4 x-7)^{2}}$

$=\frac{7(4 x-7)-(7 x+4) \cdot 4}{(4 x-7)^{2}}$

$=\frac{28 x-49-28 x-16}{(4 x-7)^{2}}$

$=\frac{-65}{(4 x-7)^{2}}$

(iii) $\frac{x^{2}+1}{x+2}$

Sol :

(iv) $\frac{x^{4}+1}{x^{2}+1}$

Sol :

(v) $\frac{x^{2}+3 x-9}{x^{2}-9 x+3}$

Sol :

माना $y=\frac{x^{2}+3 x-9}{x^{2}-9 x+3}$

Differentiating w.r.t x

$\frac{d y}{dx}=\frac{(2 x+3)\left(x^{2}-9 x+3\right)-\left(x^{2}+3 x-9\right)(2 x-9)}{\left(x^{2}-9 x+3\right)^{2}}$

$=\frac{2 x^{3}-18 x^{2}+6 x+3 x^{2}-27 x+9-2 x^{3}+9 x^{2}-6 x^{2}+27 x+18 x-81}{\left(x^{2}-9 x+3\right)^{2}}$

$=\frac{-12 x^{2}+24 x-72}{\left(x^{2}-9 x+3\right)^{2}}$

$=-12 \frac{\left(x^{2}-2 x+6\right)}{\left(x^{2}-9 x+3\right)^{2}}$

(vi) $\frac{x^{n}-a^{n}}{x-a}$

Sol :

माना $y=\frac{x^{n}-a^{n}}{x-a}$

Differentiating w.r.t x

$\frac{d y}{d x}=\frac{n \cdot x^{n-1}(x-a)-\left(x^{n}-a^{n}\right) \cdot 1}{(x-a)^{2}}$

$=\frac{n \cdot x^{n}-n a x^{n-1}-x^{n}+a^{n}}{(x-a)^{2}}$

$=\frac{(n-1) x^{n}-a\left(n x^{n-1}-a^{n-1}\right)}{(x-a)^{2}}$

(vii) $\frac{x^{2}-1}{x^{2}+7 x+1}$

Sol :

(viii) $\frac{x^{2}+3 x+1}{x^{2}-x+1}$

Sol :

(ix) $\frac{1}{p x^{2}+q x+r}$

Sol :

माना $y=\frac{1}{px^{2}+q x+r}$

Differentiating w.r.t x

$\frac{d y}{d x}=\frac{0 \cdot\left(p x^{2}+qx+r\right)-1 \cdot \frac{d\left(p x^{2}+qx+r\right)}{dx}}{(px^2+qx+r)^2}$

$=\frac{-(2 p x+q)}{\left(p x^{2}+q x+r\right)^{2}}$

(x) $\frac{x-a}{x-b}$,जहाँ a तथा b अचर है।

where a and b are constant

Sol :

माना $y=\frac{x-a}{x-b}$

Differentiating w.r.t x

$\frac{d y}{dx}=\frac{1 \cdot(x-b)-(x-a) \cdot 1}{(x-b)^{2}}$

$=\frac{x-b-x+a}{(x-b)^{2}}$

$=\frac{a-b}{(x-b)^{2}}$

(xi) $\frac{a x+b}{c x+d}$

Sol :

माना $y=\frac{a x+b}{c x+d}$

Differentiating w.r.t x

$\frac{d y}{d x}=\frac{a \cdot(c x+d)-(a x+b) \cdot c}{(cx+d)^{2}}$

$=\frac{a c x+a d-acx-b c}{(c x+d)^{2}}$

$=\frac{a d-b c}{(c x+d)^{2}}$

(xii) $\frac{a x+b}{p x^{2}+q x+r}$

Sol :

माना $y=\frac{a x+b}{ax^{2}+q x+r}$

Differentiating w.r.t x

$\frac{d{y}}{d x}=\frac{a\left(p x^{2}+q x+r\right)-(a x+b)(2 p x+q)}{\left(p x^{2}+q{x+r}\right)^{2}}$

$=\frac{a p x^{2}+a qx+a r-2 a p r^{2}-a qx-2bpx-bq}{\left(p x^{2}+q x+r\right)^{2}}$

$=-\frac{a p x^{2}-2 b{p x}+a r-b{q}}{\left(p x^{2}+q x+r\right)^{2}}$

(xiv) $\frac{2}{x+1}-\frac{x^{2}}{3 x-1}$

Sol :

माना $y=\frac{2}{x+1}-\frac{x^{2}}{3 x-1}$

Differentiating w.r.t x

$\frac{d y}{dx}=\frac{d\left(\frac{2}{x+1}\right)}{d x}-\frac{d\left(\frac{x^{2}}{3 x-1}\right)}{dx}$

$=\frac{0 \cdot(x+1)-2 \times 1}{(x+1)^{2}}-\frac{2 x \cdot(3 x-1)-x^{2} \cdot 3}{(3 x-1)^{2}}$

$=\frac{-2}{(x+1)^{2}}-\frac{6 x^{2}-2 x-3 x^{2}}{(3 x-1)^{2}}$

$=\frac{-2}{(x+1)^{2}}-\frac{3 x^{2}-2 x}{(3 x-1)^{2}}$

$=\frac{-2}{(x+1)^{2}}-\frac{x(3 x-2)}{(3 x-1)^{2}}$

Question 12

$\frac{d y}{d x}$ निकालें यदि (find $\frac{d y}{d x}$ if)

(i) $y=\frac{\sin x}{x}$

Sol :

$y=\frac{\sin x}{x}$

Differentiating w.r.t x

$\frac{d y}{dx}=\frac{\cos x \cdot(x)-\sin x \times 1}{x^{2}}$

$=\frac{x \cos x-\sin x}{x^{2}}$

(ii) $y=\frac{1-\cos x}{1+\cos x}$

Sol :

$y=\frac{1-\cos x}{1+\cos x}$

Differentiating w.r.t x

$\frac{d y}{d x}=\frac{-(-\sin x) \cdot(1+\cos x)-(1-\cos x)(-\sin x)}{(1+\cos x)^{2}}$

$=\frac{\sin x+\sin x \cos x+\sin x-\sin x \cos x}{(1+\cos x)^{2}}$

$=\frac{2 \sin x}{(1+\cos x)^{2}}$

(iii) $y=\frac{1+\tan x}{1-\tan x}$

Sol :

(iv) $y=\frac{x^{2}+\sec x}{1+\tan x}$

Sol :

$y=\frac{x^{2}+\operatorname{sec} x}{1+\tan x}$

Differentiating w.r.t x

$\frac{d y}{d x}=\frac{(2 x+\sec x \tan x)(1+\tan x)-\left(x^{2}+\sec 2\right) \sec ^{2} x}{(1+\tan x)^{2}}$

$=\frac{b c \cos x+b d \cos ^{2} x+a d \sin x+bd \sin ^{2} x}{(c+d \cos x)^{2}}$

(v) $\frac{\cos x}{1+\sin x}$

Sol :

(vi) $\frac{a+b \sin x}{c+d \cos x}$

Sol :

$y=\frac{a+b \sin x}{c+d \cos x}$

Differentiating w.r.t x

$\frac{d y}{d x}=\frac{b \cos x(c+d \cos x)-(a+\operatorname{bin} x)(-d \sin x)}{(c+d \cos x)^{2}}$

$=\frac{b c \cos x+a d \sin x+b d\left(\cos ^{2} x+\sin ^{2} x\right)}{(c+d \cos x)^{2}}$

$=\frac{b c \cos x+a d \sin x+b d}{(c+d \cos x)^{2}}$

(vii) $\frac{x+\cos x}{\tan x}$

Sol :

(viii) $\frac{\sin x-\cos x}{\sin x+\cos x}$

Sol :

$y=\frac{\sin x-\cos x}{\sin x+\cos x}$

Differentiating w.r.t x

$\frac{d y}{d x}=\frac{(\cos x+\sin x)(\sin x+\cos x)-(\sin x-\cos x)(\cos x-\sin x)}{(\sin x+\cos x)^{2}}$

$=\frac{(\sin x+\cos x)^{2}+(\sin x-\cos x)^{2}}{(\sin x+\cos x)^{2}}$

$=\frac{2\left(\sin ^{2} x+\cos ^{2} x\right)}{(\sin x+\cos x)^{2}}$

$=\frac{2}{(\sin x+\cos )^{2}}$

(ix) $\frac{\sin (x+a)}{\cos x}$

Sol :

$y=\frac{\sin (x+a)}{\cos x}=\frac{\sin x \cdot \cos a+\cos x \cdot \sin a}{\cos x}$

$y=\frac{\cos a \cdot \sin x+\sin a \cdot \cos x}{\cos x}$

Differentiating w.r.t x

$\frac{d y}{dx}=\frac{(\cos a \cdot \cos x-\sin a \sin x) \cdot \cos x-(\cos a \sin x-\sin a \cos x)(-\sin x)}{(\cos x)^{2}}$

$=\frac{\cos a \cos ^{2} x-\sin a \sin x \cos x+\cos a \sin ^{2} x+\sin a.\sin x.\cos x}{(\cos x)^{2}}$

$=\frac{\cos a\left(\cos ^{2} x+\sin ^{2} x\right)}{\cos ^{2} x}=\frac{\cos a}{\cos ^2 x}$

(x) $\frac{4 x+5 \sin x}{3 x+7 \cos x}$

Sol :

$y=\frac{4 x+5 \sin x}{3 x+7 \cos x}$

Differentiating w.r.t x

$\frac{d y}{dx}=\frac{(4+5 \cos x) \cdot(3 x+7 \cos x)-(4 x+5\sin x)\left(3-7\sin x\right)}{(3 x+7 \cos x)^{2}}$

$=\frac{12x+28\cos x+15x\cos x+35\cos^2 x-12x+28x\sin x-15\sin x+35 \sin^2 x}{(3x+7\cos x)^2}$

$=\frac{35+15 x \cos x+28 x \sin x+28 \cos x-15\sin x}{(3 x+7 \cos x)^{2}}$

(xi) $\frac{x^{5}-\cos x}{\sin x}$

Sol :

(xii) $\frac{x^{2} \cos \frac{\pi}{4}}{\sin x}$

Sol :

$y=\cos \frac{\pi}{4} \cdot \frac{x^{2}}{\sin x}$

$y=\frac{1}{\sqrt{2}} \cdot \frac{x^{2}}{\sin x}$

Differentiating w.r.t x

$\frac{d y}{d x}=\frac{1}{\sqrt{2}}\left[\frac{2 x \sin x-x^{2} \cdot \cos x}{\operatorname{sin}^{2} x}\right]$

$=\frac{x(2\sin x-x \cos x)}{\sqrt{2} \sin ^{2} x}$

Question 13

निम्नलिखित का x के सापेक्ष अवकलन करें 

[Differentiate the following w.r.t. to x] :

(i) $\frac{x \cos x}{1+x^{2}}$

Sol :

माना $y=\frac{x \cos x}{1+x^{2}}$

Differentiating w.r.t x

$\frac{d y}{d x}=\frac{\frac{d(x \cdot \cos x)}{d x} \cdot\left(1+x^{2}\right)-x \cos x \cdot d\left(\frac{1+x^{2}}{2}\right)}{\left(1+x^{2}\right)^{2}}$

$=\frac{[1 \cdot \cos x+x(-\sin x)]\left(1+x^{2}\right)-x \cos x \cdot 2 x}{\left(1+x^{2}\right)^{2}}$

$=\frac{\left(1+x^{2}\right)(\cos x-x \sin x)-2 x^{2} \cos x}{\left(1+x^{2}\right)^{2}}$

$=\frac{\left(1+x^{2}\right) \cos x-x\left(1+x^{2}\right) \sin x-2 x^{2} \cos x}{\left(1+x^{2}\right)^{2}}$

$=\frac{\left(1+x^{2}-2 x^{2}\right) \cos x-x\left(1+x^{2}\right) \sin x}{\left(1+x^{2}\right)^{2}}$

$=\frac{\left(1-x^{2}\right) \cos x-x\left(1+x^{2}\right) \sin x}{(1+x^2)^2}$

(ii) $\frac{x^{2} \sin x}{1+\tan x}$

Sol :

माना $y=\frac{x^{2} \sin x}{1+\tan x}$

Differentiating w.r.t x

$\frac{d y}{d x}=\frac{\frac{d\left(x^{2} \sin x\right)}{d x}(1+\tan x)-x^{2} \sin x \cdot \frac{d(1+\tan x)}{d x}}{(1+\tan x)^{2}}$

$=\frac{\left[2 x \sin x+x^{2} \cos x\right](1+\tan x)-x^{2} \sin x \sec ^{2} x}{(1+\tan x)^{2}}$

(iii) $\frac{x \sin x}{\sin x+\cos x}$

Sol :

(iv) $x^{2} \sec x+\frac{x^{2}}{1+\sin x}$

Sol :

माना $y=x^{2} \sec x+\frac{x^{2}}{1+\sin x}$

Differentiating w.r.t x

$\frac{d y}{dx}=\frac{d\left(x^{2} \cdot \sec x\right)}{d x}+\frac{d\left(\frac{x^{2}}{1+\sin x}\right)}{d x}$

$=2 x \sec x+x^{2} \sec x \tan x+\frac{2 x(1+\sin x)-x^{2} \cos x}{(1+\sin x)^{2}}$

Question 14

निम्नलिखित का x के सापेक्ष अवकलन करें 

[Differentiate the following w.r.t to x] :

(i) $\frac{\sec x-1}{\sec x+1}$

Sol :

(ii) $\frac{\tan x-\cot x}{\tan x+\cot x}$

Sol :

माना $y=\frac{\tan x-\cot x}{\tan x+\cot x}$

$y=\frac{\frac{\sin x}{\cos x}-\frac{\cos x}{\sin x}}{\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}}$

$y=\frac{\frac{\sin ^{2} x-\cos ^{2} x}{\cos x \sin x}}{\frac{\sin ^{2} x+\cos^{2} x}{\cos x \sin x}}$

$y=\sin ^{2} x-\cos ^{2} x$

y=(sin x-cos x)(sin x+cos x)

Differentiating w.r.t x

$\frac{d y}{d x}=$(cos x+sin x)(sin x+cos x)+(sin x-cos x)(cos x-sin x)

$=(\sin x+\cos x)^{2}-(\sin x-\cos x)^{2}$

=4sin x.cos x

=2.2sin x.cos x

=2sin 2x

(iii) $\frac{a x^{2}+b x+c}{p x^{2}+q x+r}$

Sol :

(iv) $\frac{x^{2}+\sin x}{x \cos x}$

Sol :

माना $y=\frac{x^{2}+\sin x}{x \cos x}$

Differentiating w.r.t x

$\frac{d y}{dx}=\frac{(2 x+\cos x) \cdot x \cos x-\left(x^{2}+\sin x\right)(1. \cos x+x(-\sin x)]}{(x \cos x)^{2}}$

$=\frac{2 x^{2} \cos x+x \cos ^{2} x-x^{2} \cos x+x^{3} \sin x-\sin x \cos x+x \sin^{2} x}{x^{2} \cos ^{2} x}$

$=\frac{x^{2} \cos x+x+x^{3} \sin x-\sin x \cos x}{x^{2} \cos ^{2} x}$

(v) $\frac{\sin x+x \cos x}{x \sin x-\cos x}$

Sol :

माना $y=\frac{\sin x+x \cos x}{x \sin x-\cos x}$

Differentiating w.r.t x

$\frac{d y}{d x}=\frac{[\cos x+1 \cdot \cos x+x(-\sin x)] \cdot(x \sin x-\cos x)-(\sin x+x \cos x)[1.\sin x+x\cos x -(-\sin x)]}{(x \sin x-\cos x)^{2}}$

$=\frac{(2 \cos x-x \sin x)(x \sin x-\cos x)-(\sin x+x \cos x)(2\sin x+x\cos x)}{ \left.(x \sin x-\cos x)^{2}\right.}$

$=\frac{2 x \sin x \cos x-2 \cos ^{2} x-x^{2} \sin ^{2} x+x \sin x \cos x-2 \sin^2 x-x \sin x \cos x+2 x \sin x \cos x+x^{2} \cos ^{2} x}{(x\sin x-\cos x)^2}$

$=\frac{-2\left(\cos ^{2} x+\sin ^{2} x\right)-x^{2}\left(\sin ^{2} x+\cos ^{2} x\right)}{(x \sin x-\cos x)^{2}}$

$=\frac{-2-x^{2}}{(x-\sin x-\cos x)^{2}}$

$=-\frac{\left(2+x^{2}\right)}{(x \sin x-\cos x)^{2}}$

Question 15

यदि (If) $y=\frac{x}{x+a}$, साबित करें कि (prove that) $x \frac{d y}{d x}=y(1-y)$.

Sol :

$y=\frac{x}{x+a}$

Differentiating w.r.t x

$\frac{d y}{dx}=\frac{1 \cdot(x+a)-x(1)}{(x+a)^{2}}$

$\frac{d y}{d}=\frac{(x+a)-x}{(x+a)^{2}}$

$\frac{d y}{dx}=\frac{x+a}{(x+a)^{2}}-\frac{x}{(x+a)^2}$

$\frac{d y}{dx}=\frac{1}{x+a}-\frac{x}{(x+a)^{2}}$

$x \frac{d y}{dx}=\frac{2}{x+a}-\frac{x^{2}}{(x+a)^2}$

$x\frac{dy}{dx}=y-y^{2}$

$x \frac{d y}{dx}=y(1-y)$

Question 16

अवकलन के भागफल नियम से साबित करें कि 

(Using quotient rule for differentiate, prove that)

(i) $\frac{d}{d x}(\tan x)=\sec ^{2} x$

Sol :

$\frac{d}{d x}(\tan x)=\frac{d\left(\frac{\sin x}{\cos x}\right)}{dx}$

$=\frac{\cos x \cdot \cos x-\sin x(-\sin x)}{\cos ^{2} x}$

$=\frac{\cos ^{2} x+\sin ^{2} x}{\cos ^{2} x}$

$=\frac{1}{\cos ^{2} x}=-\sec^{2} x$

(ii) $\frac{d^{\prime}}{d x}(\operatorname{cosec} x)=-\operatorname{cosec} x \cot x$

Sol :

Question 17

निम्नलिखित फलनों का x के सापेक्ष अवकल गुणांक निकालें। 

[Find the d.c. of the following functions w.r. to x] :

(i) $(a x+b)^{n}$

Sol :

माना $y=(a x+b)^{n}$

Differentiating w.r.t. x

$\frac{d y}{d x}=\frac{d(a x+b)^{n}}{d(a x+b)} \times \frac{d(ax+b)}{d x}$

$=n(a x+b)^{n-1} \cdot a$

$=n a \cdot(a x+b)^{n-1}$

(ii) $\frac{x}{\cos ^{n} x}$

Sol :

माना $y=\frac{x}{\cos ^{n} x}$

Differentiating w.r.t. x

$\frac{d y}{d x}=\frac{1 \cdot \cos ^{n} x-x \cdot \frac{d\left(\cos ^{n} x\right)}{d(\cos x)} \times \frac{d(\cos x)}{dx}}{\cos ^{2 n} x}$

$=\frac{\cos ^{n} x-x \cdot n \cos ^{n-1} x \cdot(-\sin x)}{\cos ^{2 n} x}$

$=\frac{\cos ^{n-1} x(\cos x+n x \sin x)}{\cos ^{2 n} x}$

$=\frac{\cos x+n x \sin x}{\cos ^{n+1} x}$