Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 4 - Calculating the Derivative - 4.5 Derivatives of Logarithmic Functions - 4.5 Exercises: 20

Answer

\[{y^,} = \frac{{6{x^3}\ln x - 2{x^3} + 2}}{{4x{{\ln }^2}x}}\]

Work Step by Step

\[\begin{gathered} y = \frac{{{x^3} - 1}}{{2\ln x}} \hfill \\ Differentiate \hfill \\ {y^,} = \,\,{\left[ {\frac{{{x^3} - 1}}{{2\ln x}}} \right]^,} \hfill \\ Use\,\,the\,\,quotient\,\,rule \hfill \\ {y^,} = \frac{{\,\left( {2\ln x} \right)\,{{\left( {{x^3} - 1} \right)}^,} - \,\left( {{x^3} - 1} \right)\,\left( {2\ln x} \right)}}{{\,{{\left( {2\ln x} \right)}^2}}} \hfill \\ Then \hfill \\ {y^,} = \frac{{\,\left( {2\ln x} \right)\,\left( {3{x^2}} \right) - \,\left( {{x^3} - 1} \right)\,\left( 2 \right)\,\left( {\frac{1}{x}} \right)}}{{\,{{\left( {2\ln x} \right)}^2}}} \hfill \\ Simplifying \hfill \\ {y^,} = \frac{{6{x^2}\ln x - \frac{{2{x^3} - 2}}{x}}}{{4{{\ln }^2}x}} \hfill \\ {y^,} = \frac{{6{x^2}\ln x - \frac{{2{x^3} - 2}}{x}}}{{4{{\ln }^2}x}} \hfill \\ {y^,} = \frac{{6{x^3}\ln x - 2{x^3} + 2}}{{4x{{\ln }^2}x}} \hfill \\ \end{gathered} \]
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