Question

Differentiate the following functions with respect to x
(without using first principle) :

$\sf i) \: \sqrt{sin \: x} \\ \\ ii) \: \sf sin \sqrt{x}$
​

Answer 1

$\huge\mathfrak{solution}$

In both questions we use ,

Chain rule .

$\star \boxed{ \bold{ \frac{d}{dx} f(g(x)) = f' (x) \frac{d}{dx} g(x)}}$

$1) \sqrt{sin \: x}$

Let

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

Differentiate w.r.t X both sides ,

$\rightarrow \: \frac{dy}{dx} = \frac{d}{dx} ( \sqrt{sin \: x} ) \\ \\ \rightarrow \: \frac{dy}{dx} = \frac{1}{2 \sqrt{sin \: x} } \frac{d}{dx} (sin \: x) \\ \\ \rightarrow \: \frac{dy}{dx} = \frac{1}{2 \sqrt{sin \: x} } (cosx) \\ \\ \rightarrow \: \frac{dy}{dx} = \frac{cos \: x}{2 \sqrt{sin \: x} }$

$2) \: sin \: \sqrt{x}$

Let

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

Differentiate w.r.t X both sides,

$\star \frac{dy}{dx} = \frac{d}{dx} (sin \: \sqrt{x} ) \\ \\ \star \frac{dy}{dx} = cos \: \sqrt{x} \frac{d}{dx} ( \sqrt{x} ) \\ \\ \star \frac{dy}{dx} = cos \: \sqrt{x} ( \frac{1}{2 \sqrt{x} } ) \\ \\ \star \: \frac{dy}{dx} = \frac{cos \: \sqrt{x} }{2 \sqrt{x} }$

Formula used:

$\red \star \boxed{ \bold{ \frac{d}{dx} ( \sqrt{x} ) = \frac{1}{2 \sqrt{x} } }}$

$\green\star \boxed{\bold{ \frac{d}{dx} (sin \: x) = cos \: x}}$

Answer 2

$\: \: \: \: \: \: \: \: \: \: \: \huge {\red{\boxed{ \overline{ \underline{ \mid\mathfrak{An}{\mathrm{sw}{ \sf{er}} \colon\mid}}}}}}$

(1) $\sf \: \sqrt{\sin \: x}$

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$\sf{let \: y \: = \: \sqrt{ \sin x}} \\ \\ \implies {\sf{\dfrac{dy}{dx} \: = \: \dfrac{d}{dx} \big( \sqrt{\sin x} \big) }} \\ \\ \implies {\sf{\dfrac{dy}{dx} \: = \: \dfrac{1}{ 2 \sqrt{ \sin x}} \dfrac{d}{dx} \big( \sin x \big)}} \\ \\ \implies {\sf{\dfrac{dy}{dx} \: = \: \dfrac{1}{2 \sqrt{\sin x}} \big( \cos x \big) }} \\ \\ \implies {\sf{\dfrac{dy}{dx} \: = \: \dfrac{\cos x}{2 \sqrt{\sin x}}}}$

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(2) $\sf {\sin \sqrt{x}}$

$\sf{let \: y \: = \: \sin \sqrt{x}} \\ \\ \implies {\sf{\dfrac{dy}{dx} = \dfrac{d}{dx} \big( \sin \: \sqrt{x} \big)}} \\ \\ \implies {\sf{ \dfrac{dy}{dx} \: = \: \cos \: \sqrt{x} \dfrac{d}{dx} \big( \sqrt{x} \big)}} \\ \\ \implies {\sf{ \dfrac{dy}{dx} = \cos \: \sqrt{x} \bigg( \dfrac{1}{2 \sqrt{x} } \bigg)}} \\ \\ \implies {\sf{ \dfrac{dy}{dx} = \dfrac{\cos \: \sqrt{x} }{2 \sqrt{x} }}}$