Paul's Online Notes
Home / Calculus I / Derivatives / Logarithmic Differentiation
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 3.13 : Logarithmic Differentiation

2. Use logarithmic differentiation to find the first derivative of $$\displaystyle y = \frac{{\sin \left( {3z + {z^2}} \right)}}{{{{\left( {6 - {z^4}} \right)}^3}}}$$.

Show All Steps Hide All Steps

Start Solution

Take the logarithm of both sides and do a little simplifying.

\begin{align*}\ln \left( y \right) & = \ln \left[ {\frac{{\sin \left( {3z + {z^2}} \right)}}{{{{\left( {6 - {z^4}} \right)}^3}}}} \right] = \ln \left[ {\sin \left( {3z + {z^2}} \right)} \right] - \ln \left[ {{{\left( {6 - {z^4}} \right)}^3}} \right]\\ & = \ln \left[ {\sin \left( {3z + {z^2}} \right)} \right] - 3\ln \left[ {6 - {z^4}} \right]\end{align*} Show Step 2

Use implicit differentiation to differentiate both sides with respect to $$z$$.

$\frac{{y'}}{y} = \frac{{\left( {3 + 2z} \right)\cos \left( {3z + {z^2}} \right)}}{{\sin \left( {3z + {z^2}} \right)}} - 3\left[ {\frac{{ - 4{z^3}}}{{6 - {z^4}}}} \right] = \left( {3 + 2z} \right)\cot \left( {3z + {z^2}} \right) + \frac{{12{z^3}}}{{6 - {z^4}}}$ Show Step 3

Finally, solve for the derivative and plug in the equation for $$y$$ .

$y' = y\left[ {\left( {3 + 2z} \right)\cot \left( {3z + {z^2}} \right) + \frac{{12{z^3}}}{{6 - {z^4}}}} \right] = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{\sin \left( {3z + {z^2}} \right)}}{{{{\left( {6 - {z^4}} \right)}^3}}}\left[ {\left( {3 + 2z} \right)\cot \left( {3z + {z^2}} \right) + \frac{{12{z^3}}}{{6 - {z^4}}}} \right]}}$