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 SolutionTake 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 2Use 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 3Finally, solve for the derivative and plug in the equation for \(y\) .
\[y' = y\left[ {\left( {3 + 2z} \right)\cot \left( {3z + {z^2}} \right) + \frac{{12{z^3}}}{{6 - {z^4}}}} \right] = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{\sin \left( {3z + {z^2}} \right)}}{{{{\left( {6 - {z^4}} \right)}^3}}}\left[ {\left( {3 + 2z} \right)\cot \left( {3z + {z^2}} \right) + \frac{{12{z^3}}}{{6 - {z^4}}}} \right]}}\]