Continuity and Differentiability — Class 12 Maths Solution

Let $y = {\sin ^m}x \cdot {\cos ^n}x$
therefore,$\frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {{{(\sin x)}^m} \cdot {{(\cos x)}^n}} \right]$

$= {(\sin x)^m} \cdot \frac{d}{{dx}}{(\cos x)^n} + {(\cos x)^n} \cdot \frac{d}{{dx}}{(\sin x)^m}$

$= {(\sin x)^m} \cdot n{(\cos x)^{n - 1}} \cdot \frac{d}{{dx}}\cos x + {(\cos x)^n}m{(\sin x)^{m - 1}} \cdot \frac{d}{{dx}}\sin x$
$= {(\sin x)^m} \cdot n{(\cos x)^{n - 1}}( - \sin x) + {(\cos x)^n} \cdot m{(\sin x)^{m - 1}}\cos x$

$= - n{\sin ^m}x \cdot {\cos ^{n - 1}}x \cdot (\sin x) + m{\cos ^n}x \cdot {\sin ^{m - 1}}x \cdot \cos x$

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

$= - n \cdot {\sin ^m}x \cdot {\cos ^n}x \cdot \tan x + m{\sin ^m}x \cdot {\cos ^n}x \cdot \cot x$

$= {\sin ^m}x \cdot {\cos ^n}x[ - n\tan x + m\cot x]$

$\frac{dy}{dx} = = {\sin ^m}x \cdot {\cos ^n}x[ - n\tan x + m\cot x]$