Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 3 - Derivatives - Review Exercises - Page 234: 52


$$\frac{{f'\left( {\sqrt {g\left( x \right)} } \right)\left( {g'\left( x \right)} \right)}}{{2\sqrt {g\left( x \right)} }}$$

Work Step by Step

$$\eqalign{ & \frac{d}{{dx}}f\left( {\sqrt {g\left( x \right)} } \right) \cr & {\text{Let }}f\left( x \right){\text{ and }}g\left( x \right){\text{ differentiable functions with }}g\left( x \right) \geqslant 0 \cr & {\text{use the chain rule}} \cr & \frac{d}{{dx}}f\left( {\sqrt {g\left( x \right)} } \right) = \frac{d}{{dx}}\left[ {f\left( {\sqrt {g\left( x \right)} } \right)} \right]\frac{d}{{dx}}\left[ {\sqrt {g\left( x \right)} } \right] \cr & {\text{rewrite }}\sqrt {g\left( x \right)} {\text{ as }}{\left( {g\left( x \right)} \right)^{1/2}} \cr & \frac{d}{{dx}}f\left( {\sqrt {g\left( x \right)} } \right) = \frac{d}{{dx}}\left[ {f\left( {\sqrt {g\left( x \right)} } \right)} \right]\frac{d}{{dx}}\left[ {{{\left( {g\left( x \right)} \right)}^{1/2}}} \right] \cr & {\text{use the chain rule again}} \cr & \frac{d}{{dx}}f\left( {\sqrt {g\left( x \right)} } \right) = \frac{d}{{dx}}\left[ {f\left( {\sqrt {g\left( x \right)} } \right)} \right]\left[ {\frac{1}{2}{{\left( {g\left( x \right)} \right)}^{ - 1/2}}} \right]\frac{d}{{dx}}\left[ {g\left( x \right)} \right] \cr & {\text{simplifying}} \cr & \frac{d}{{dx}}f\left( {\sqrt {g\left( x \right)} } \right) = f'\left( {\sqrt {g\left( x \right)} } \right)\left[ {\frac{1}{2}{{\left( {g\left( x \right)} \right)}^{ - 1/2}}} \right]\left( {g'\left( x \right)} \right) \cr & \frac{d}{{dx}}f\left( {\sqrt {g\left( x \right)} } \right) = f'\left( {\sqrt {g\left( x \right)} } \right)\left[ {\frac{1}{{2\sqrt {g\left( x \right)} }}} \right]\left( {g'\left( x \right)} \right) \cr & \frac{d}{{dx}}f\left( {\sqrt {g\left( x \right)} } \right) = \frac{{f'\left( {\sqrt {g\left( x \right)} } \right)\left( {g'\left( x \right)} \right)}}{{2\sqrt {g\left( x \right)} }} \cr} $$
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