# Differentiate the following functions:

Question:

Differentiate the following functions:

(i) If $y=6 x^{5}-4 x^{4}-2 x^{2}+5 x-9$, find $\frac{d y}{d x}$ at $x=-1$

(ii) If $y=(\sin x+\tan x)$, find $\frac{d y}{d x}$ at $x=\frac{\pi}{3}$.

(iii) If $y=\frac{(2-3 \cos x)}{\sin x}$, find $\frac{d y}{d x}$ at $x=\frac{\pi}{4}$.

Solution:

Formulae:

$\frac{d}{d x} x^{n}=n x^{n-1}$

$\frac{d}{d x} \cot x=-\operatorname{cosec}^{2} x$

$\frac{d}{d x} \operatorname{cosec} x=-\operatorname{cosec} x \cot x$

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

$\frac{d}{d x} \sin x=\cos x$

(i) If $y=6 x^{5}-4 x^{4}-2 x^{2}+5 x-9$, find $\frac{d y}{d x}$ at $x=-1$

Differentiating with respect to $x$,

$\frac{d}{d x}\left(6 x^{5}-4 x^{4}-2 x^{2}+5 x-9\right)$

$=30 x^{4}-16 x^{3}-4 x+5$

substituing $\mathrm{x}=-1$

$\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) x=-1=30(-1)^{4}-16(-1)^{3}-4(-1)+5$

$=30+16+4+5$

$=55$

(ii) If $y=(\sin x+\tan x)$, find $\frac{d y}{d x}$ at $x=\frac{\pi}{3}$.

Differentiating with respect to $x$,

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

Substituting $x=\frac{\pi}{3}$

$\left(\frac{d y}{d x}\right) x=\pi / 3=\cos \frac{\pi}{3}+\sec \frac{2 \pi}{3}$

$=\frac{1}{2}+4$

$=\frac{5}{2}$

(iii) If $\mathrm{y}=\frac{(2-3 \cos \mathrm{x})}{\sin \mathrm{x}}$, find $\frac{\mathrm{dy}}{\mathrm{dx}}$ at $\mathrm{x}=\frac{\pi}{4}$

Differentiating with respect to $x$,

$\frac{d}{d x}(2 \operatorname{cosec} x-3 \cot x)=2(-\operatorname{cosec} x \cot x)-3\left(-\operatorname{cosec}^{2} x\right)$

Substituting $\mathrm{x}=\frac{\pi}{4}$

$\left(\frac{d y}{d x}\right) x=\pi / 4=2\left(-\operatorname{cosec} \frac{\pi}{4} \cot \frac{\pi}{4}\right)-3\left(-\operatorname{cosec} \frac{2 \pi}{4}\right)$

$=-2 \times \sqrt{2}+3 \times 2$

$=6-2 \times \sqrt{2}$