26A06-ExampleOfChainRule.tex

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\pmtitle{example of chain rule}
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\begin{document}
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 
$$h'(x)=f'(g(x))g'(x).$$

Since
$$f'(x)=\frac{1}{2\sqrt{x}},\quad\mbox{and}\quad
g'(x)=\cos(x),$$
we have by chain rule
$$h'(x) = \left(\frac{1}{2\sqrt{\sin x}}\right)\cos x=\frac{\cos
  x}{2\sqrt{\sin 
x}}$$

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
\begin{align*}
\frac{dz}{dx} &= \frac{1}{2\sqrt{y}}\, \cos(x) \\
&= \frac{1}{2}\,\frac{\cos(x)}{\sqrt{\sin(x)}}
\end{align*}
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\end{document}