# Foreshadowing the Chain Rule

I assigned another very easy but good problem this week. It was simple enough, but it gave a hint of things to come.

Use the Product Rule to find the derivative of ${{\left( f\left( x \right) \right)}^{2}}$.

Since we have not yet discussed the Chain Rule, the Product Rule was the only way to go.

$\frac{d}{dx}{{\left( f \right)}^{2}}=\frac{d}{dx}\left( f\cdot f \right)=f\cdot {f}'+{f}'\cdot f=2f\cdot f'$

And likewise for higher powers:

$\frac{d}{dx}{{f}^{3}}=\frac{d}{dx}\left( f\cdot f\cdot f \right)=f\cdot f\cdot {f}'+f\cdot {f}'\cdot f+{f}'\cdot f\cdot f=3{{f}^{2}}{f}'$

If you just look at the answer, it is not clear where the ${f}'$  comes from. But the result foreshadows the Chain Rule.

Then we used the new formula to differentiate a few expressions such as ${{\left( 4x+7 \right)}^{2}}$ and ${{\sin }^{2}}\left( x \right)$ and a few others.

Regarding the Chain Rule: I have always been a proponent of the Rule of Four, but I have never seen a good graphical explanation of the Chain Rule. (If someone has one, PLEASE send it to me – I’ll share it.)

Here is a rough verbal explanation that might help a little.

Consider the graph of $y=\sin \left( x \right)$. On the interval $[0,2\pi ]$ it goes through all its value in order once – from 0 to 1 to 0 to -1 and back to zero. Now consider the graph of $y=\sin \left( 3x \right)$. On the interval $\left[ 0,\tfrac{2\pi }{3} \right]$ it goes through all the same values in one-third of the time. Therefore, it must go through them three times as fast. So the rate of change of $y=\sin \left( 3x \right)$ between 0 and $\tfrac{2\pi }{3}$ must be three times the rate of change of $y=\sin \left( x \right)$. So the rate of change of  must be $3\cos \left( 3x \right)$. Of course this rate of change is the slope and the derivative.

This site uses Akismet to reduce spam. Learn how your comment data is processed.