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

21. Differentiate \(\displaystyle K\left( x \right) = \frac{{1 + {{\bf{e}}^{ - 2x}}}}{{x + \tan \left( {12x} \right)}}\) .

Hint : Don’t forget the Product and Quotient Rule. Sometimes, in the process of doing the Product or Quotient Rule you’ll need to use the Chain Rule when differentiating one or both of the terms in the product or quotient.
Show Solution

For this problem we’ll need to do the Quotient Rule to start off the derivative. In the process we’ll need to use the Chain Rule when we differentiate both the numerator and the denominator.

The derivative is then,

\[\require{bbox} \bbox[2pt,border:1px solid black]{{K'\left( x \right) = \frac{{ - 2{{\bf{e}}^{ - 2x}}\left( {x + \tan \left( {12x} \right)} \right) - \left( {1 + {{\bf{e}}^{ - 2x}}} \right)\left( {1 + 12{{\sec }^2}\left( {12x} \right)} \right)}}{{{{\left( {x + \tan \left( {12x} \right)} \right)}^2}}}}}\]