Paul's Online Notes
Paul's Online Notes
Home / Calculus I / Derivatives / Chain Rule
Show Mobile Notice Show All Notes Hide All Notes
Mobile Notice
You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

Section 3.9 : Chain Rule

3. Differentiate \(y = \sqrt[3]{{1 - 8z}}\) .

Hint : Recall that with Chain Rule problems you need to identify the “inside” and “outside” functions and then apply the chain rule.
Show Solution

For this problem, after converting the root to a fractional exponent, the outside function is (hopefully) clearly the exponent of \(\frac{1}{3}\) while the inside function is the polynomial that is being raised to the power (or the polynomial inside the root – depending upon how you want to think about it). The derivative is then,

\[y = {\left( {1 - 8z} \right)^{\frac{1}{3}}}\hspace{0.25in}\hspace{0.25in} \Rightarrow \hspace{0.25in}\,\,\,\,\,\,\,\,\,\,\frac{{dy}}{{dz}} = \frac{1}{3}{\left( {1 - 8z} \right)^{ - \,\,\frac{2}{3}}}\left( { - 8} \right) = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \frac{8}{3}{{\left( {1 - 8z} \right)}^{ - \,\,\frac{2}{3}}}}}\]