Question

$\small {\mathrm{ If \: \: y = \sqrt{x} \sin x +\sin\sqrt{x} } }$
$\small {\mathrm{ Show \: \: that \: \: \dfrac{dy}{dx} =\dfrac{1}{2 \sqrt{x} }(\sin x + 2x \cos x + \cos \sqrt{x}) } }$
Explanation needed..!!
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Answer 1

ANSWER:

Given:

$\:\:\bullet\:\:y=\sqrt{x}\sin x+\sin\sqrt{x}$

To Prove:

$\:\:\bullet\:\:\dfrac{dy}{dx}=\dfrac{1}{2\sqrt{x}}\left(\sin x+2x\cos x+\cos\sqrt{x}\right)$

Proof:

We are given that,

$\implies y=\sqrt{x}\sin x+\sin\sqrt{x}$

Differentiating both sides, w.r.t. x,

$\implies\dfrac{dy}{dx}=\dfrac{d}{dx}\bigg(\sqrt{x}\sin x+\sin\sqrt{x}\bigg)$

We know that,

$\hookrightarrow\dfrac{d}{dx}\left(f(x)+g(x)\bigg)=\dfrac{d}{dx}\left(f(x)\bigg)+\dfrac{d}{dx}\left(g(x)\bigg)$

So,

$\implies\dfrac{dy}{dx}=\dfrac{d}{dx}\bigg(\sqrt{x}\sin x+\sin\sqrt{x}\bigg)$

$\implies\dfrac{dy}{dx}=\dfrac{d}{dx}\bigg(\sqrt{x}\sin x\bigg)+\dfrac{d}{dx}\bigg(\sin\sqrt{x}\bigg)$

We know that, by product rule,

$\hookrightarrow [f(x)g(x)]'=f(x)g'(x)+g(x)f'(x)$

And, by chain rule,

$\hookrightarrow [f(g(x))]' = f'(g(x))\times g'(x)$

So, applying product rule in first term, and chain rule in both,

$\implies\dfrac{dy}{dx}=\dfrac{d}{dx}\bigg(\sqrt{x}\sin x\bigg)+\dfrac{d}{dx}\bigg(\sin\sqrt{x}\bigg)$

$\implies\dfrac{dy}{dx}=\left(\sqrt{x}\dfrac{d}{dx}\left(\sin x\right)+\sin x\dfrac{d}{dx}\left(\sqrt{x}\right)\right)+\left(\dfrac{d}{dx}\left(\sin\sqrt{x}\right)\times\dfrac{d}{dx}\left(\sqrt{x}\right)\right)$

We know that,

$\hookrightarrow\dfrac{d}{dx}\sin x=\cos x$

$\hookrightarrow\dfrac{d}{dx}\sqrt{x}= \dfrac{d}{dx}x^{\frac{1}{2}}=\dfrac{1\times x^{\frac{1}{2}-1}}{2} =\dfrac{1\times x^{\frac{-1}{2}}}{2}=\dfrac{1}{2\sqrt{x}}$

And,

$\hookrightarrow\dfrac{d}{dx} x=1$

So,

$\implies\dfrac{dy}{dx}=\left(\sqrt{x}\dfrac{d}{dx}\left(\sin x\right)+\sin x\dfrac{d}{dx}\left(\sqrt{x}\right)\right)+\left(\dfrac{d}{dx}\left(\sin\sqrt{x}\right)\times\dfrac{d}{dx}\left(\sqrt{x}\right)\right)$

$\implies\dfrac{dy}{dx}=\left((\sqrt{x}\times\cos x)+\left(\sin x\times\dfrac{1}{2\sqrt{x}}\right)\right)+\left(\cos\sqrt{x}\times\dfrac{1}{2\sqrt{x}}\right)$

$\implies\dfrac{dy}{dx}=\sqrt{x}\cos x+\dfrac{\sin x}{2\sqrt{x}}+\dfrac{\cos\sqrt{x}}{2\sqrt{x}}\right)$

Taking LCM,

$\implies\dfrac{dy}{dx}=\dfrac{(\sqrt{x}\cos x)(2\sqrt{x})}{2\sqrt{x}}+\dfrac{\sin x}{2\sqrt{x}}+\dfrac{\cos\sqrt{x}}{2\sqrt{x}}\right)$

$\implies\dfrac{dy}{dx}=\dfrac{2x\cos x}{2\sqrt{x}}+\dfrac{\sin x}{2\sqrt{x}}+\dfrac{\cos\sqrt{x}}{2\sqrt{x}}\right)$

Taking 1/2√x common,

$\implies\dfrac{dy}{dx}=\dfrac{2x\cos x}{2\sqrt{x}}+\dfrac{\sin x}{2\sqrt{x}}+\dfrac{\cos\sqrt{x}}{2\sqrt{x}}\right)$

$\implies\dfrac{dy}{dx}=\dfrac{1}{2\sqrt{x}}\bigg(2x\cos x+\sin x+\cos\sqrt{x}\bigg)$

Rearranging the terms,

$\implies\bf\dfrac{dy}{dx}=\dfrac{1}{2\sqrt{x}}\bigg(sin x+2x\,cos x+cos\sqrt{x}\bigg)$

Hence Proved!!

Answer 2

Hey User ,

pls, refer to the attachment done by me.