Question

q1.​find dy/dx,if y=cos√sinx

Answer 1

Step-by-step explanation:

Differentiation

Given, y = cos (√sinx)

$</p><p>\begin{gathered}y = \cos( \sqrt{sinx} ) \\ \\ \frac{dy}{dx} = \frac{d}{dx} (\cos( \sqrt{sinx} ) \: \\ \\ \frac{dy}{dx} = - sin( \sqrt{sinx} ) \times \frac{ d}{dx} ( \sqrt{sinx} ) \\ \\ \frac{dy}{dx} = - sin( \sqrt{sinx} ) \times \frac{ 1}{2 ( \sqrt{sinx} ) } \times \frac{d}{dx}(sinx) \\ \\ \frac{dy}{dx} = - sin( \sqrt{sinx} ) \times \frac{ 1}{2 ( \sqrt{sinx} ) } \times cosx \\ \\ \frac{dy}{dx} = \dfrac{ - cosx. \sin( \sqrt{ sinx } ) }{2( \sqrt{sinx} )} \end{gathered}$

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

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