Calculus: Early Transcendentals (2nd Edition)

Published by Pearson

Chapter 6 - Applications of Integration - 6.8 Logarithmic and Exponential - 6.8 Exercises: 35

Answer

$$h'\left( x \right) = {2^{{x^2} + 1}}x\left( {\ln 2} \right)$$

Work Step by Step

\eqalign{ & h\left( x \right) = {2^{\left( {{x^2}} \right)}} \cr & {\text{evaluate the derivative}} \cr & h'\left( x \right) = \frac{d}{{dx}}\left( {{2^{\left( {{x^2}} \right)}}} \right) \cr & {\text{use }}\frac{d}{{dx}}\left( {{a^{u\left( x \right)}}} \right) = {a^{u\left( x \right)}}\left( {\ln a} \right)u'\left( x \right) \cr & h'\left( x \right) = {2^{\left( {{x^2}} \right)}}\left( {\ln 2} \right)\frac{d}{{dx}}\left( {{x^2}} \right) \cr & h'\left( x \right) = {2^{\left( {{x^2}} \right)}}\left( {\ln 2} \right)\left( {2x} \right) \cr & {\text{simplify}} \cr & h'\left( x \right) = {2^{{x^2} + 1}}x\left( {\ln 2} \right) \cr}

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