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}\quad\mbox{and}\quad g(x)=\sin(x).

Then chain rule says that

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

Since

f^{\prime}(x)=\frac{1}{2\sqrt{x}},\quad\mbox{and}\quad 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 the following appearance. First we describe the functional relation as

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

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

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

We then have

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

and hence the chain rule gives

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