Question

pls friend help.me ​

Answer 1

Answer:

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Answer 2

$\huge \mathfrak{solution}$

To Differentiate:

$y = ({ {x}^{2} + 1 })^{ \sin \: x}$

Take Log both sides ,we get

$\star \: log(y) = log( ({ {x}^{2} + 1})^{ \sin \: x} ) \\ \\ \star \: log(y) = \sin x \: \: log( {x}^{2} + 1)$

Now Differentiate w.r.t X

$\star \: \frac{d}{dx} log(y) = \frac{d}{dx} ( \sin x \: log( {x}^{2} + 1 ) ) \\ \\ \star \: \frac{1}{y} \frac{dy}{dx} = \sin x \: \frac{d}{dx} log( {x}^{2} + 1 ) + log( {x}^{2} + 1 ) \frac{d}{dx} ( \sin x) \\ \\ \star \: \frac{1}{y} \frac{dy}{dx} = \sin x \frac{1}{ {x}^{2} + 1 } (2x + 0) + log( {x}^{2} + 1) ( \cos x) \\ \\ \star \: \frac{1}{y} \frac{dy}{dx} = \frac{2x \sin x}{ {x}^{2} + 1 } + \cos x \: log( {x}^{2} + 1 ) \\ \\ \star \: \frac{dy}{dx} = y( \frac{2x \: \sin x}{ {x}^{2} + 1 } + \cos x \: log( {x}^{2} + 1 ) ) \\ \\ \star \: \frac{dy}{dx} = {( {x}^{2} + 1) }^{ \sin x} ( \frac{2x \sin x}{ {x}^{2} + 1 } + \cos x \: log( {x}^{2} + 1) )$

Formula

$\star \ \bold{ \frac{d}{dx} u.v = u \: \frac{dv}{dx} + v \: \frac{du}{dx}} \\ \\ \star \bold{ \frac{d}{dx} \sin \: x = \cos x}$