Section 3.13 : Logarithmic Differentiation
3. Use logarithmic differentiation to find the first derivative of \(\displaystyle h\left( t \right) = \frac{{\sqrt {5t + 8} \,\,\,\sqrt[3]{{1 - 9\cos \left( {4t} \right)}}}}{{\sqrt[4]{{{t^2} + 10t}}}}\).
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Start SolutionTake the logarithm of both sides and do a little simplifying.
\[\begin{align*}\ln \left[ {h\left( t \right)} \right] & = \ln \left[ {\frac{{\sqrt {5t + 8} \,\,\,\sqrt[3]{{1 - 9\cos \left( {4t} \right)}}}}{{\sqrt[4]{{{t^2} + 10t}}}}} \right]\\ & = \ln \left[ {\sqrt {5t + 8} \,\,\,\sqrt[3]{{1 - 9\cos \left( {4t} \right)}}} \right] - \ln \left[ {\sqrt[4]{{{t^2} + 10t}}} \right]\\ & = \ln \left[ {{{\left( {5t + 8} \right)}^{\frac{1}{2}}}} \right] + \ln \left[ {{{\left( {1 - 9\cos \left( {4t} \right)} \right)}^{\frac{1}{3}}}} \right] - \ln \left[ {{{\left( {{t^2} + 10t} \right)}^{\frac{1}{4}}}} \right]\\ & = {\textstyle{1 \over 2}}\ln \left( {5t + 8} \right) + {\textstyle{1 \over 3}}\ln \left( {1 - 9\cos \left( {4t} \right)} \right) - {\textstyle{1 \over 4}}\ln\left( {{t^2} + 10t} \right)\end{align*}\]Note that the logarithm simplification work was a little complicated for this problem, but if you know your logarithm properties you should be okay with that.
Show Step 2Use implicit differentiation to differentiate both sides with respect to \(t\).
\[\frac{{h'\left( t \right)}}{{h\left( t \right)}} = {\textstyle{1 \over 2}}\frac{5}{{5t + 8}} + {\textstyle{1 \over 3}}\frac{{36\sin \left( {4t} \right)}}{{1 - 9\cos \left( {4t} \right)}} - {\textstyle{1 \over 4}}\frac{{2t + 10}}{{{t^2} + 10t}}\] Show Step 3Finally, solve for the derivative and plug in the equation for \(h\left( t \right)\) .
\[\begin{align*}h'\left( t \right) & = h\left( t \right)\left[ {\frac{{{\textstyle{5 \over 2}}}}{{5t + 8}} + \frac{{12\sin \left( {4t} \right)}}{{1 - 9\cos \left( {4t} \right)}} - \frac{{{\textstyle{1 \over 2}}t + {\textstyle{5 \over 2}}}}{{{t^2} + 10t}}} \right]\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{\sqrt {5t + 8} \,\,\,\sqrt[3]{{1 - 9\cos \left( {4t} \right)}}}}{{\sqrt[4]{{{t^2} + 10t}}}}\left[ {\frac{{{\textstyle{5 \over 2}}}}{{5t + 8}} + \frac{{12\sin \left( {4t} \right)}}{{1 - 9\cos \left( {4t} \right)}} - \frac{{{\textstyle{1 \over 2}}t + {\textstyle{5 \over 2}}}}{{{t^2} + 10t}}} \right]}}\end{align*}\]