example of chain rule

Suppose we wanted to differentiate

$$h(x)=\sqrt{\sin (x)}.$$

Here, $h(x)$ is given by the composition

$$h(x)=f(g(x)),$$

where

$$f(x)=\sqrt{x}\mathit{\hspace{1em}}\text{and}\mathit{\hspace{1em}}g(x)=\sin (x).$$

Then chain rulesays that

$$h^{\prime }(x)=f^{\prime }(g(x))g^{\prime }(x).$$

Since

$$f^{\prime }(x)=\frac{1}{2\sqrt{x}},\text{and}\mathit{\hspace{1em}}{g}^{\prime }(x)=\cos (x),$$

we have by chain rule

$$h^{\prime }(x)=\left(\frac{1}{2\sqrt{\sin x}}\right)\cos x=\frac{\cos x}{2\sqrt{\sin x}}$$

Using the Leibniz formalism, the above calculation would have thefollowing appearance. First we describe the functional relation as

$$z=\sqrt{\sin (x)}.$$

Next, we introduce an auxiliary variable $y$, and write

$$z=\sqrt{y},y=\sin (x).$$

We then have

$$\frac{dz}{dy}=\frac{1}{2\sqrt{y}},\frac{dy}{dx}=\cos (x),$$

and hence the chain rule gives

${\displaystyle \frac{dz}{dx}}$$={\displaystyle \frac{1}{2\sqrt{y}}}\cos (x)$

$={\displaystyle \frac{1}{2}}{\displaystyle \frac{\cos (x)}{\sqrt{\sin (x)}}}$