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: 18

Answer

\[{y^,} = \frac{{\frac{{ - 6x + 2}}{x} + 6\ln x}}{{\,{{\left( {3x - 1} \right)}^2}}}\]

Work Step by Step

\[\begin{gathered} y = \frac{{ - 2\ln x}}{{3x - 1}} \hfill \\ Differentiate \hfill \\ {y^,} = \,\,\left[ {\frac{{ - 2\ln x}}{{3x - 1}}} \right] \hfill \\ Use\,\,the\,\,quotient\,\,rule \hfill \\ {y^,} = \frac{{\,\left( {3x - 1} \right)\,{{\left( { - 2\ln x} \right)}^,}\,\left( { - 2\ln x} \right)\,{{\left( {3x - 1} \right)}^,}}}{{\,{{\left( {3x - 1} \right)}^2}}} \hfill \\ \operatorname{Re} call\,\,that\,\,\frac{d}{{dx}}\,\,\left[ {\ln x} \right] = \frac{1}{x} \hfill \\ {y^,} = \frac{{\,\left( {3x - 1} \right)\,\left( { - 2} \right)\,\left( {\frac{1}{x}} \right) + 2\ln x\,\left( 3 \right)}}{{\,{{\left( {3x - 1} \right)}^2}}} \hfill \\ Multiplying \hfill \\ {y^,} = \frac{{\frac{{ - 6x + 2}}{x} + 6\ln x}}{{\,{{\left( {3x - 1} \right)}^2}}} \hfill \\ \hfill \\ \end{gathered} \]
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