## Calculus with Applications (10th Edition)

Published by Pearson

# Chapter 4 - Calculating the Derivative - 4.4 Derivatives of Exponential Functions - 4.4 Exercises - Page 232: 24

#### Answer

${f^,}\,\left( t \right) = \,\left( {6t{e^{{t^2}}} + 15} \right)\,{\left( {{e^{{t^2}}} + 5t} \right)^2}$

#### Work Step by Step

$\begin{gathered} f\,\left( t \right) = \,{\left( {{e^{{t^2}}} + 5t} \right)^3} \hfill \\ Use\,\,the\,\,generalized\,\,power\,\,rule \hfill \\ \frac{d}{{dx}}\,\,{\left[ {g\,\left( x \right)} \right]^n} = n\,\,{\left[ {g\,\left( x \right)} \right]^{n - 1}}{g^,}\,\left( x \right) \hfill \\ Then \hfill \\ {f^,}\,\left( t \right) = 3{\left( {{e^{{t^2}}} + 5t} \right)^{3 - 1}}\,{\left( {{e^{{t^2}}} + 5t} \right)^,} \hfill \\ Differentiate\,\,{e^{{t^2}}} + 5t \hfill \\ {f^,}\,\left( t \right) = 3{\left( {{e^{{t^2}}} + 5t} \right)^2}\,\left( {2t{e^{{t^2}}} + 5} \right) \hfill \\ Multiplying \hfill \\ {f^,}\,\left( t \right) = \,\left( {6t{e^{{t^2}}} + 15} \right)\,{\left( {{e^{{t^2}}} + 5t} \right)^2} \hfill \\ \end{gathered}$

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