# example of chain rule

Suppose we wanted to differentiate

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

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

 $\displaystyle\frac{dz}{dx}$ $\displaystyle=\frac{1}{2\sqrt{y}}\,\cos(x)$ $\displaystyle=\frac{1}{2}\,\frac{\cos(x)}{\sqrt{\sin(x)}}$
Title example of chain rule ExampleOfChainRule 2013-03-22 12:35:32 2013-03-22 12:35:32 rmilson (146) rmilson (146) 4 rmilson (146) Example msc 26A06