Question

$\sf Differentiate\:following \:function \: w.r.t.\: x$


$\sf {e}^{{x}^{2}{sin}^{2}+ {cos}^{2}x}$



$\sf Answer\:is \: {e}^{{x}^{2}{sin}^{2}+ {cos}^{2}x} (({x}^{2}-1)sin 2x + 2x {sin}^{2}x)$


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Answer 1

Solution

Derivative of y is $\sf {e}^{{x}^{2}{sin}^{2}+ {cos}^{2}x}[sin2x(x^2 - 1) + 2x sin^2 x]$

Given

$\sf \: y = \ {e}^{{x}^{2}{sin}^{2}+ {cos}^{2}x}$

To finD

The derivative of y

Differentiating y w.r.t to x,we get :

$\sf \: y' = (\sf {e}^{{x}^{2}{sin}^{2}+ {cos}^{2}x}) \times \dfrac{d( {x}^{2} {sin}^{2}x + {cos}^{2}x) }{dx} \\ \\ \longrightarrow \: \sf \: y' =( {e}^{{x}^{2}{sin}^{2}+ {cos}^{2}x } ) \times \dfrac{d(cos {}^{2}x) }{dx} + \dfrac{d(x {}^{2} {sin}^{2}x) }{dx}$

Using Product Rule for cos²x and x²sin²x

$\sf a = cos {}^{2} x = (cos \: x)(cos \: x) = uv \\ \\ \implies \: \sf \: a' = \dfrac{d(cos \: x)}{dx}(cos \: x )+ \frac{d(cos \: x)}{dx} (cos \ x ) \\ \\ \implies \: \sf \: a' = ( - \: sin \: x)cos \: x + ( - sin \: x)(cos \: x) \\ \\ \implies \: \sf \: a' = - 2 \: sin \: x \: cos \: x \\ \\ \implies \: \sf \: a' = - sin2x$

Also,

$\sf \: b' = {x}^{2} {sin}^{2}x = (x^2)(sin^2x) = uv \\ \\ \implies \sf \: b' = \frac{d(x^2)}{dx}( \: sin^2 \: x) + \frac{d(sin^2 x)}{dx} (x^2 )\\ \\ \implies \: \sf \: b' = 2x(sin^2 x) + x^2 sin2x \\ \\ \implies \: \sf \: b' = x^2 sin2x + 2x sin^2x$

Now,

$\longrightarrow \sf \: y' = {e}^{{x}^{2}{sin}^{2}+ {cos}^{2}x} \: [(x^2 sin2x + 2x sin^2x \: - sin2x] \\ \\ \huge{\longrightarrow \boxed{\boxed{ \sf \: y' =( {e}^{{x}^{2}{sin}^{2}+ {cos}^{2}x} )[ sin2x(x^2 - 1) + 2x sin^2 x]}}}$

Note

• Derivative of e^x is e^x

• Derivative of cos x is - sin x

• Derivative of sin x is cos x

• Product Rule : u'v + v'u

• Derivative of sin²x is sin2x

• Derivative of cos²x is - sin2x

Answer 2

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

$\bigstar\sf{\:Differentiate\:following \:function \: w.r.t.\:x}$

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

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

We Know,

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

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

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

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

$\mapsto\pink{\sf{\:2\sin x \cos x \:=\:\sin 2x}}$

_____________________

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

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

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

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

$\:\:\:\:\green{\sf{\:\left[\dfrac{duv}{dx}\:=\:u\dfrac{dv}{dx}+v\dfrac{du}{dx}\right]}}$

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

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

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

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

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

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

That's Answer