Question

find dy/dx,if y=cos√sinx
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Answer 1

Answer:

y=cosVsinx diffrentiate w.r.to x, dy/dx=d/dx (cosVsinx) =-sinVsinx.d/dx (Vsinx) =-sinVsinx.1/2Vsinx.d/dx (sinx) dy/dx=-sinVsinx.1/2Vsinx.cosx

Answer 2

Differentiation

Given, y = cos (√sinx)

$y = \cos( \sqrt{sinx} ) \\ \\ \frac{dy}{dx} = \frac{d}{dx} (\cos( \sqrt{sinx} ) \: \\ \\ \frac{dy}{dx} = - sin( \sqrt{sinx} ) \times \frac{ d}{dx} ( \sqrt{sinx} ) \\ \\ \frac{dy}{dx} = - sin( \sqrt{sinx} ) \times \frac{ 1}{2 ( \sqrt{sinx} ) } \times \frac{d}{dx}(sinx) \\ \\ \frac{dy}{dx} = - sin( \sqrt{sinx} ) \times \frac{ 1}{2 ( \sqrt{sinx} ) } \times cosx \\ \\ \frac{dy}{dx} = \dfrac{ - cosx. \sin( \sqrt{ sinx } ) }{2( \sqrt{sinx} )}$

We have used the chain rule of differentiation.

$\frac{dy}{dx} = \frac{dy}{dt} \times \frac{dt}{dx}$

Differentiation of sine function.

$\frac{d}{dx} (sinx) = cosx$

Differentiation of cosine function

$\frac{d}{dx} (cosx) = - sinx$

Differentiation of square function

$\frac{d}{dx} ( \sqrt{x} ) = \frac{1}{2 \sqrt{x} }$