Question

if y=sin√cosx then find dy/dx

Answer 1

This derivative in this question will be found using the chain rule,

$\frac{dy}{dx} =\frac{du}{dx} \cdot \frac{dy}{du}$.

To simplify the problem, we fist should make the substitution,

$u = \sqrt{cos(x)}$.

After this substitution, the problem then becomes that of finding the derivative of ,

$y=sin(u)$.

Next, we calculate the two derivatives that go on the right hand side of the chain rule:

$\frac{dy}{du}= cos(u).$

and

$\frac{du}{dx} =-\frac{1}{2} \cdot sin(x)\cdot\frac{1}{\sqrt{cos(x)}}$

$\frac{du}{dx} =-\frac{sin(x)}{2\sqrt{cos(x)}}$

The derivative of the function is then

$\frac{dy}{dx} =\frac{du}{dx} \cdot\frac{dy}{du}$

$\frac{du}{dx} =-\frac{sin(x)}{2\sqrt{cos(x)}} \cdot cos(u)$

$\frac{du}{dx} =-\frac{sin(x)}{2\sqrt{cos(x)}} \cdot cos(\sqrt{cos(x)})$

$\frac{du}{dx} =-\frac{sin(x)\cdot cos(\sqrt{cos(x)})}{2\sqrt{cos(x)}}$