Question

If y=square root of sin x, find dy/dx​

Answer 1

✦Question :-

$\sf \red{ if \: y = \sqrt{ \sin x} } \green{ \: , \: then \: find \: \frac{dy}{dx} }$

✦Answer :-

$\implies\sf \boxed{ \sf \green{\leadsto } \: \red{ \frac{dy}{dx} }= \blue{ \frac{ \cos x}{2 \sqrt{ \sin x} }} } \:$

❏ Solution :-

We have ,

$\leadsto \sf \red{ y =} \purple{ \sqrt{ \sin x} } \\ \\ \bf squaring \: on \: both \: sides \: \\ \\ \leadsto \sf \green{{y}^{2} = } \pink{ {( \sqrt{ \sin x} )}^{2} } \\ \\ \leadsto \sf \purple{ {y}^{2} } = \orange{ \sin x} \\ \\ \bf differentiate \: with \: respect \: to \: x \\ \\ \leadsto \sf \orange{ \frac{d}{dx} \: {y}^{2} }= \green{\frac{d}{dx} \sin x} \\ \\ \because \: \sf \blue{ \frac{d}{x} \: } {x}^{n} = \red{ n \: {x}^{n - 1} } \: \: , \green{ \frac{d}{dx} \sin x = }\cos x \\ \\ \therefore \\ \leadsto \sf \: \orange{2y \: \frac{dy}{dx} }= \cos x \\ \\ \leadsto \sf \red{ \frac{dy}{dx} } = \green{\frac{ \cos x}{2y} } \\ \\ \because \sf \: y = \sqrt{ \sin x} \\ \\ \sf \boxed{ \sf \green{\leadsto } \: \red{ \frac{dy}{dx} }= \blue{ \frac{ \cos x}{2 \sqrt{ \sin x} }} }$

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Hope it helps you .

Answer 2

If $y=\sqrt{sinx}$ then $\frac{dy}{dx}$ is   $\frac{dy}{dx} = \frac{cosx}{2\sqrt{sinx}}$

Given:

y=square root of sin x that is $y=\sqrt{sinx}$.

To find:

$\frac{dy}{dx}$

Solution:

$y=\sqrt{sinx}$

Squaring both sides we get,

$y^{2} =(\sqrt{sinx} )^2$

$y^{2} =sinx$

Now differentiate above with respect to x we get:

$2y\frac{dy}{dx} = cosx$       ...(using identity $\frac{d}{dx} x^{n} =n x^{n-1}$ and $\frac{d}{dx} sinx=cosx$)

$\frac{dy}{dx} = \frac{cosx}{2y}$

As $y=\sqrt{sinx}$  we will put this in the above equation,

$\frac{dy}{dx} = \frac{cosx}{2\sqrt{sinx}}$

Hence, If $y=\sqrt{sinx}$ then $\frac{dy}{dx}$ is   $\frac{dy}{dx} = \frac{cosx}{2\sqrt{sinx}}$.

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