Calculus with Applications (10th Edition)

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

Chapter 4 - Calculating the Derivative - 4.4 Derivatives of Exponential Functions - 4.4 Exercises: 18

Answer

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

Work Step by Step

\[\begin{gathered} y = \frac{{{e^x}}}{{2x + 1}} \hfill \\ Differentiate\,\,using\,\,the\,\,quotient\,\,rule \hfill \\ \,\,{\left[ {\frac{u}{v}} \right]^,} = \frac{{v{u^,} - u{v^,}}}{{{v^2}}} \hfill \\ Then \hfill \\ {y^,} = \frac{{\,\left( {2x + 1} \right)\,{{\left( {{e^x}} \right)}^,} - {e^x}\,{{\left( {2x + 1} \right)}^,}}}{{\,{{\left( {2x + 1} \right)}^2}}} \hfill \\ Compute\,\,the\,\,derivatives \hfill \\ {y^,} = \frac{{\,\left( {2x + 1} \right)\,\left( {{e^x}} \right) - {e^x}\,\,{{\left( 2 \right)}^,}}}{{\,{{\left( {2x + 1} \right)}^2}}} \hfill \\ Multiply \hfill \\ {y^,} = \frac{{2x{e^x} + {e^x} - 2{e^x}}}{{\,{{\left( {2x + 1} \right)}^2}}} \hfill \\ {y^,} = \frac{{2x{e^x} - {e^x}}}{{\,{{\left( {2x + 1} \right)}^2}}} \hfill \\ \end{gathered} \]
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