The Chain Rule


\begin{displaymath}\begin{array}{cc}\Big(f \circ g(x)\Big)' = g'(x) f'(g(x))\\
...
...aystyle{\frac{dy}{dx} = \frac{dy}{du}
\frac{du}{dx}}\end{array}\end{displaymath}

As a motivation for the chain rule, consider the function

f(x) = (1+x2)10.

Since f(x) is a polynomial function, we know from previous pages that f'(x) exists. Naturally one may ask for an explicit formula for it. One tedious way to do this is to develop (1+x2)10 using the Binomial Formula and then take the derivative. Of course, it is possible to do this, but it won't be much fun. But what if we have to deal with (1+x2)100! Then I hope you agree that the Binomial Formula is not the way to go anymore.

So what do we do? The answer is given by the Chain Rule. Before we discuss the Chain Rule formula, let us give another example.

Example. Let us find the derivative of $f(x) = \sin(2x)$. One way to do that is through some trigonometric identities. Indeed, we have

\begin{displaymath}\sin(2x) = 2\sin(x) \cos(x) \cdot\end{displaymath}

So we will use the product formula to get

\begin{displaymath}\Big(\sin(2x)\Big)' = 2 \Big(\sin'(x) \cos(x) + \sin(x) \cos'(x)\Big)\end{displaymath}

which implies

\begin{displaymath}\Big(\sin(2x)\Big)' = 2 \Big(\cos^2(x) - \sin^2(x)\Big)\cdot\end{displaymath}

Using the trigonometric formula $\cos(2x) = \cos^2(x) -
\sin^2(x)$, we get

\begin{displaymath}\Big(\sin(2x)\Big)' = 2 \cos(2x)\cdot\end{displaymath}

Once this is done, you may ask about the derivative of $\sin(5x)$? The answer can be found using similar trigonometric identities, but the calculations are not as easy as before. Again we will see how the Chain Rule formula will answer this question in an elegant way.

In both examples, the function f(x) may be viewed as:

\begin{displaymath}f(x) = h\Big(g(x)\Big)\end{displaymath}

where g(x) = 1+x2 and h(x) = x10 in the first example, and $h(x) = \sin(x)$ and g(x) = 2x in the second. We say that f(x) is the composition of the functions g(x) and h(x) and write

\begin{displaymath}f(x) = h \circ g(x).\end{displaymath}

The derivative of the composition is given by the formula

\begin{displaymath}\Big(f \circ g(x)\Big)' = g'(x) f'(g(x)).\end{displaymath}

Another way to write this formula is

\begin{displaymath}\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx}\end{displaymath}

where $y = f \circ g(x) = f(u)$ and u = g(x). This second formulation (due to Leibniz) is easier to remember and is the formulation used almost exclusively by physicists.

Example. Let us find the derivative of

\begin{displaymath}f(x) = (1+x^2)^{100}\cdot\end{displaymath}

We have $f(x) = h\circ g(x)$, where g(x) = 1+x2 and h(x) = x100. Then the Chain rule implies that f'(x) exists, which we knew since it is a polynomial function, and

\begin{displaymath}f'(x) = 2x \cdot \Big[100(1+x^2)^{99}\Big] = 200 x (1+x^2)^{99}\;.\end{displaymath}

Example. Let us find the derivative of

\begin{displaymath}f(x) = \sin(5x)\cdot\end{displaymath}

We have $f(x) = h\circ g(x)$, where g(x) = 5x and $h(x) = \sin(x)$. Then the Chain rule implies that f'(x) exists and

\begin{displaymath}f'(x) = 5\cdot \Big[\cos(5x)\Big]= 5 \cos(5x)\cdot\end{displaymath}

In fact, this is a particular case of the following formula

\begin{displaymath}\Big[ f(ax+b)\Big]' = a f'(ax+b) \cdot\end{displaymath}

The following formulas come in handy in many areas of techniques of integration.


\begin{displaymath}\begin{array}{llllllll}
\displaystyle \frac{d}{dx}(u^n) &=& n...
...}(e^u) &=& e^u \displaystyle \frac{du}{dx}\\
&&\\
\end{array}\end{displaymath}

More formulas for derivatives can be found in our section of tables.


Exercise 1. Find the derivative of

\begin{displaymath}f(t) = \left(t^2 - \frac{2}{t^3} \right)^2\;.\end{displaymath}

Answer.


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Mohamed A. Khamsi

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