Power Rule Implies Chain Rule

Having developed the Product Rule d\left( uv \right)=u{v}'+{u}'v and the Power Rule \frac{d}{dx}{{x}^{n}}=n{{x}^{n-1}} for derivatives in your class, you can explore similar rules for the product of more than two functions and suddenly the Chain Rule will appear.

For three functions use the associative property of multiplication with the rule above:

d\left( uvw \right)=d\left( \left( uv \right)w \right)=u\cdot v\cdot dw+w\cdot d(uv)=u\cdot v\cdot dw+w\left( udv+vdu \right)

So expanding with a slight change in notation:

d\left( uvw \right)=uv{w}'+u{v}'w+u'vw

For four factors there is a similar result:

d\left( uvwz \right)=uvw{z}'+uv{w}'z+u{v}'wz+{u}'vwz

Exercise: Let {{f}_{i}} for i=1,2,3,...,n be functions. Write a general formula for the derivative of the product {{f}_{1}}{{f}_{2}}{{f}_{3}}\cdots {{f}_{n}} as above and in sigma notation


d\left( {{f}_{1}}{{f}_{2}}{{f}_{3}}\cdots {{f}_{n}} \right)={{f}_{1}}{{f}_{2}}{{f}_{3}}\cdots {{{f}'}_{n}}+{{f}_{1}}{{f}_{2}}{{{f}'}_{3}}\cdots {{f}_{n}}+{{f}_{1}}{{{f}'}_{2}}{{f}_{3}}\cdots {{f}_{n}}+\cdots +{{{f}'}_{1}}{{f}_{2}}{{f}_{3}}\cdots {{f}_{n}}

\displaystyle d\left( {{f}_{1}}{{f}_{2}}{{f}_{3}}\cdots {{f}_{n}} \right)=\sum\limits_{i=1}^{n}{\frac{{{f}_{1}}{{f}_{2}}{{f}_{3}}\cdots {{f}_{n}}}{{{f}_{i}}}{{{{f}'}}_{i}}}

This idea may now be used  to see the Chain Rule appear. Students may guess that d{{\left( f \right)}^{4}}=4{{\left( f \right)}^{3}}, but wait there is more to it.

Write {{\left( f \right)}^{4}}=f\cdot f\cdot f\cdot f\text{ }. Then from above

d{{\left( f \right)}^{4}}=d\left( f\cdot f\cdot f\cdot f\text{ } \right)=f\cdot f\cdot f\cdot {f}'+f\cdot f\cdot {f}'\cdot f+f\cdot {f}'\cdot f\cdot f+{f}'\cdot f\cdot f\cdot f

d{{\left( f \right)}^{4}}=4{{\left( f \right)}^{3}}{f}'\text{ }

Looks just like the power rule, but there’s that “extra” {f}'. Now you are ready to explain about the Chain Rule in the next class.


Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s