Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - Review Exercises - Page 393: 40

Answer

$$h'\left( z \right) = - z{e^{ - \frac{{{z^2}}}{2}}}$$

Work Step by Step

$$\eqalign{ & h\left( z \right) = {e^{ - \frac{{{z^2}}}{2}}} \cr & {\text{Differentiate}} \cr & h'\left( z \right) = \frac{d}{{dz}}\left[ {{e^{ - \frac{{{z^2}}}{2}}}} \right] \cr & h'\left( z \right) = {e^{ - \frac{{{z^2}}}{2}}}\frac{d}{{dz}}\left[ { - \frac{{{z^2}}}{2}} \right] \cr & h'\left( z \right) = {e^{ - \frac{{{z^2}}}{2}}}\left( {\frac{{ - 2z}}{2}} \right) \cr & h'\left( z \right) = - z{e^{ - \frac{{{z^2}}}{2}}} \cr} $$
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