I have been informed that on March 7th from 6:00am to 6:00pm Central Time Lamar University will be doing some maintenance to replace a faulty UPS component and to do this they will be completely powering down their data center.
Unfortunately, this means that the site will be down during this time. I apologize for any inconvenience this might cause.
Paul
February 18, 2026
Section 3.13 : Logarithmic Differentiation
5. Find the first derivative of \(f\left( x \right) = {\left( {2x - {{\bf{e}}^{8x}}} \right)^{\sin \left( {2x} \right)}}\).
Show All Steps Hide All Steps
Start SolutionWe just need to do some logarithmic differentiation so take the logarithm of both sides and do a little simplifying.
\[\ln \left[ {f\left( x \right)} \right] = \ln \left[ {{{\left( {2x - {{\bf{e}}^{8x}}} \right)}^{\sin \left( {2x} \right)}}} \right] = \sin \left( {2x} \right)\ln \left( {2x - {{\bf{e}}^{8x}}} \right)\] Show Step 2Use implicit differentiation to differentiate both sides with respect to \(x\). Don’t forget to product rule the right side.
\[\begin{align*}\frac{{f'\left( x \right)}}{{f\left( x \right)}} & = 2\cos \left( {2x} \right)\ln \left( {2x - {{\bf{e}}^{8x}}} \right) + \sin \left( {2x} \right)\frac{{2 - 8{{\bf{e}}^{8x}}}}{{2x - {{\bf{e}}^{8x}}}}\\ & = 2\cos \left( {2x} \right)\ln \left( {2x - {{\bf{e}}^{8x}}} \right) + \sin \left( {2x} \right)\frac{{2 - 8{{\bf{e}}^{8x}}}}{{2x - {{\bf{e}}^{8x}}}}\end{align*}\] Show Step 3Finally, solve for the derivative and plug in the equation for \(f\left( x \right)\) .
\[\begin{align*}f'\left( x \right) & = f\left( x \right)\left[ {2\cos \left( {2x} \right)\ln \left( {2x - {{\bf{e}}^{8x}}} \right) + \sin \left( {2x} \right)\frac{{2 - 8{{\bf{e}}^{8x}}}}{{2x - {{\bf{e}}^{8x}}}}} \right]\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{{{\left( {2x - {{\bf{e}}^{8x}}} \right)}^{\sin \left( {2x} \right)}}\left[ {2\cos \left( {2x} \right)\ln \left( {2x - {{\bf{e}}^{8x}}} \right) + \sin \left( {2x} \right)\frac{{2 - 8{{\bf{e}}^{8x}}}}{{2x - {{\bf{e}}^{8x}}}}} \right]}}\end{align*}\]