The Chain Rule

Except for the simplest functions, a procedure known as the Chain Rule is very helpful and often necessary to find derivatives. You can start with an example such as finding the derivative of  {{\left( 2x+7 \right)}^{2}}.  Most students will expand the binomial to get 4{{x}^{2}}+28x+49 and differentiate the result to get 8x+28. They will try the same approach with {{\left( 2x+7 \right)}^{3}} and then you can hit them with {{\left( 2x+7 \right)}^{53}}.  They will see the need for a short cut at once. What to do?

The explanation runs like this. Let u\left( x \right)={{x}^{53}} and let v\left( x \right)=2x+7. Then our original expression becomes {{\left( 2x+7 \right)}^{53}}=u\left( v\left( x \right) \right) a composition of functions. The Chain Rule is used for differentiating compositions. Students must get good at recognizing compositions. The differentiation is done from the outside, working inward.  It is done in the exact opposite order than the procedure for evaluating expression. To evaluate the expression above you (1) evaluate the expression inside the parentheses and the (2) raise that result to the 53 power. To differentiate you (1) use the power rule to differentiate the 53 power of whatever is inside, this gives 53{{\left( 2x+7 \right)}^{52}}, the (2) differentiate the \left( 2x+7 \right) which give 2 and multiply the results: 53{{\left( 2x+{{7}^{52}} \right)}^{52}}(2)=106{{\left( 2x+7 \right)}^{52}}. Symbolically, this looks like {u}'\left( v\left( x \right) \right){v}'\left( x \right) or {f}'\left( g\left( x \right) \right){g}'\left( x \right). This can be extended to compositions of more than two functions:

\displaystyle \frac{d}{dx}f\left( g\left( h\left( x \right) \right) \right)={f}'\left( g\left( h\left( x \right) \right) \right){g}'\left( h\left( x \right) \right){h}'\left( x \right)

The cartoon below is from Courtney Gibbons’ great collection of math cartoons ( may help you kids remember this:

I have been looking for a way to illustrate the Chain Rule graphically, but to no avail. The closest I could come up with is this: Consider f\left( x \right)=\sin \left( 3x \right). This function takes on all the values of y=\sin \left( x \right) in order in one-third the time. (That is its period is one-third of the period of y=\sin \left( x \right). Since this is true, it must go through the values three times as fast; thus, its derivative (it’s rate of change) must be three times the derivative of the sine: {f}'\left( x \right)=3\cos \left( 3x \right).

The students will need some practice on using the Chain Rule. I suggest a number of simple (single compositions) first and then a few longer ones and maybe one or two “monsters” just for fun once they get the idea.

The Chain Rule doesn’t end with just being able to differentiate complicated expressions; it will also form the basis for implicit differentiation, finding the derivative of a function’s inverse and Related Rate problems among others things.

Finally, here is a way to develop the Chain Rule which is probably different and a little more intuitive from what you will find in your textbook. (After a suggestion by Paul Zorn on the AP Calculus EDG October 14, 2002)

Let f be a function differentiable at x=a, and let g be a function that is differentiable at x=b and such that g\left( b \right)=a. Then, near x=a we can use the local linear approximation of f and g to find  \frac{d}{dx}f\left( g\left( b \right) \right):

f\left( x \right)\approx f\left( a \right)+{f}'\left( a \right)\left( x-a \right)

f\left( g\left( x \right) \right)\approx f\left( a \right)+{f}'\left( a \right)\left( g\left( x \right)-a \right)=f\left( a \right)+{f}'\left( a \right)g\left( x \right)-a {f}'\left( a \right)

\displaystyle \frac{d}{dx}f\left( g\left( x \right) \right)=0+{f}'\left( a \right){g}'\left( x \right)-0

\displaystyle\frac{d}{dx}f\left( g\left( b \right) \right)={f}'\left( g\left( b \right) \right){g}'\left( b \right)


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