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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
1. Use logarithmic differentiation to find the first derivative of \(f\left( x \right) = {\left( {5 - 3{x^2}} \right)^7}\,\,\sqrt {6{x^2} + 8x - 12} \).
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Start SolutionTake the logarithm of both sides and do a little simplifying.
\[\begin{align*}\ln \left[ {f\left( x \right)} \right] & = \ln \left[ {{{\left( {5 - 3{x^2}} \right)}^7}\,\,\sqrt {6{x^2} + 8x - 12} } \right]\\ & = \ln \left[ {{{\left( {5 - 3{x^2}} \right)}^7}} \right] + \ln \left[ {{{\left( {6{x^2} + 8x - 12} \right)}^{{\textstyle{1 \over 2}}}}} \right]\\ & = 7\ln \left( {5 - 3{x^2}} \right) + {\textstyle{1 \over 2}}\ln \left( {6{x^2} + 8x - 12} \right)\end{align*}\] Show Step 2Use implicit differentiation to differentiate both sides with respect to \(x\).
\[\frac{{f'\left( x \right)}}{{f\left( x \right)}} = 7\frac{{ - 6x}}{{5 - 3{x^2}}} + \frac{1}{2}\frac{{12x + 8}}{{6{x^2} + 8x - 12}} = \frac{{ - 42x}}{{5 - 3{x^2}}} + \frac{{6x + 4}}{{6{x^2} + 8x - 12}}\] Show Step 3Finally, solve for the derivative and plug in the equation for \(f\left( x \right)\) .
\[\begin{align*}f'\left( x \right) & = f\left( x \right)\left[ {\frac{{ - 42x}}{{5 - 3{x^2}}} + \frac{{6x + 4}}{{6{x^2} + 8x - 12}}} \right]\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{{{\left( {5 - 3{x^2}} \right)}^7}\,\,\sqrt {6{x^2} + 8x - 12} \left[ {\frac{{ - 42x}}{{5 - 3{x^2}}} + \frac{{6x + 4}}{{6{x^2} + 8x - 12}}} \right]}}\end{align*}\]