KC Sinha Mathematics Solution Class 12 Chapter 11 अवकलन (Differentiation) Exercise 11.1 (Q5-Q8)

Exercise 11.1

Question 5

tan x⁰
Sol :
Let y=tan x⁰

$=\tan x \times \frac{\pi}{180}$

$y=\tan \frac{\pi x}{180}$

Differentiating with respect to x

$\frac{d y}{d x}=\frac{d\left(\tan \frac{\pi x}{180}\right)}{d\left(\frac{\pi x}{180}\right)} \times \frac{d\left(\frac{\pi x}{180}\right)}{d x}$

$=\sec ^{2} \frac{\pi x}{120} \times \frac{\pi}{180}$

$=\frac{\pi}{180} \cdot \sec ^{2} x^{0}$

Question 6

$\left(3 x^{2}+6 x+5\right)^{\frac{7}{2}}$
Sol :
Let y=$\left(3 x^{2}+6 x+5\right)^{\frac{7}{2}}$

Differentiating with respect to x

$\frac{d y}{dx}=\frac{d\left(3 x^{2}+6 x+5\right)^{\frac{7}{2}}}{d\left(3 x^{2}+6 x+5\right)} \times \frac{d\left(3 x^{2}+6 x+5\right)}{dx}$

$=\frac{7}{2}\left(3 x^{2}+6 x+5\right)^{5 / 2} \cdot(6 x+6)$

$=21(x+1) \cdot\left(3 x^{2}+6 x+5\right)^{5 / 2}$

Question 7

$\sqrt{5+2 x-4 x^{5}}$
Sol :
Let y=$\sqrt{5+2 x-4 x^{5}}$

Differentiating with respect to x

$\frac{dy}{dx}=\frac{d(\sqrt{5+2 x-4x^5} )}{d(5+2x-4x^5)} \times d\left(\frac{5+2x-4 x^{5}}{dx}\right)$

$=\frac{1}{2 \sqrt{5+2 x-4 x^{5}}} \times\left(2-20 x^{4}\right)$

$=\frac{2\left(1-10 x^{4}\right)}{2 \sqrt{5+2 x-4 x^{5}}}$

$=\frac{1-10 x^{4}}{\sqrt{5+2 x-4 x^5}}$

Question 8

sin(cos x³)

Sol :
Let y=sin(cos x³)

Differentiating with respect to x

$\frac{d{y}}{dx}=\frac{d\left[\sin \left(\cos x^{3}\right)\right]}{d\left(\cos x^{3}\right)} \times \frac{d\left(\cos x^{3}\right)}{d\left(x^{3}\right)} \times \frac{d\left(x^{2}\right)}{d x}$

=cos(cos x³).(-sin x³).3x²

=-3x²sinx³.cos(cosx³)