Question

$\sf Differentiate\:following \:function \: w.r.t.\: x$
$\sf {e}^{{x}^{2}{sin}^{2}+ {cos}^{2}x}$
$\sf Answer\:is \: {e}^{{x}^{2}{sin}^{2}+ {cos}^{2}x} (({x}^{2}-1)sin 2x + 2x {sin}^{2}x)$ $\sf Content Quality Answer Required$​

Answer 1

Question :-

$</p><p>\bigstar\sf{\:Differentiate\:following \:function \: w.r.t.\:x}$

$</p><p>\mapsto\orange{\sf{\:{e}^{({x}^{2}{sin}^{2} x+ {cos}^{2} x)}}}$

$\Large{\underline{\underline{\mathfrak{\orange{\bf{Solution}}}}}}$

We Know,

$\mapsto\boxed{\boxed{\pink{\sf{\:\dfrac{de^y}{dx}\:=\:e^y\dfrac{dy}{dx}}}}}$

$\mapsto\boxed{\boxed{\pink{\sf{\:\dfrac{d\sin x}{dx}\:=\:\cos x}}}}$

$</p><p>\mapsto\boxed{\boxed{\pink{\sf{\:\dfrac{d\cos x}{dx}\:=\:-\sin x}}}}$

$\mapsto\boxed{\boxed{\pink{\sf{\:\dfrac{dx^n}{dx}\:=\:nx^{n-1}}}}}$

$</p><p>\mapsto\pink{\sf{\:2\sin x \cos x \:=\:\sin 2x}}$

_____________________

$\Large{\underline{\underline{\mathfrak{\pink{\bf{Explanation}}}}}}$

$\mapsto\orange{\sf{\:\dfrac{d{e}^{(x^2\sin^2 x + \cos^2 x)}}{dx}}}$

$</p><p>\mapsto\sf{\:({e}^{x^2\sin^2 x+\cos^2 x})\:[\dfrac{d(x^2\sin^2 x + \cos^2 x)}{dx}]}$

$\mapsto\sf{\:({e}^{x^2\sin^2 x+\cos^2 x})[\dfrac{d(x^2\sin^2 x)}{dx}\:+\:\dfrac{d(\cos^2 x)}{dx}]}$

$</p><p>\:\:\:\:\green{\sf{\:[\dfrac{duv}{dx}\:=\:u\dfrac{dv}{dx}+v\dfrac{du}{dx}]}}$

$\mapsto\sf{\:({e}^{x^2\sin^2 x+\cos^2 x})\:[x^2\dfrac{d(\sin^2 x)}{dx}+\sin^2 x\:\dfrac{dx^2}{dx}\:+\:2\cos x\times\dfrac{d(\cos x)}{dx}]}$

$\mapsto\sf{\:({e}^{x^2\sin^2 x+\cos^2 x})[x^2\times2\sin x\times\dfrac{d(\sin x)}{dx}+2x\sin^2 x\:+\:2\cos x\times(-\sin x)]}$

$\mapsto\sf{\:({e}^{x^2\sin^2 x+\cos^2 x})[x^2\times2\sin x\:\cos x\:+\:2x\:\sin^2 x\:+\:2\cos x\:(-\sin x)]}$

$\:\:\:\:\:\green{\sf{\:(2\sin x \cos x \:=\:\sin 2x)}}$

$\mapsto\sf{\:({e}^{x^2\sin^2 x+\cos^2 x})[x^2\:\sin 2x\:+\:2x\:\sin^2 x\:\:-\:\sin 2x]}$

$\mapsto\sf{\:({e}^{x^2\sin^2 x+\cos^2 x})[\sin 2x\:(x^2-1)\:+\:2x\:\sin^2 x]}$

Hence proved !!

Answer 2

hope it helps you..........✌️✌️