Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 5: Integrals - Section 5.4 - The Fundamental Theorem of Calculus - Exercises 5.4 - Page 287: 42


$$\frac{{dy}}{{dx}} = 2{x^2}\sin {x^6} + \int_2^{{x^2}} {\sin \left( {{t^3}} \right)} dt $$

Work Step by Step

$$\eqalign{ & y = x\int_2^{{x^2}} {\sin \left( {{t^3}} \right)} dt \cr & {\text{differentiate both sides with respect to }}x \cr & \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {x\int_2^{{x^2}} {\sin \left( {{t^3}} \right)} dt} \right] \cr & {\text{use the product rule}} \cr & \frac{{dy}}{{dx}} = x\frac{d}{{dx}}\left[ {\int_2^{{x^2}} {\sin \left( {{t^3}} \right)} dt} \right] + \int_2^{{x^2}} {\sin \left( {{t^3}} \right)} dt\frac{d}{{dx}}\left[ x \right] \cr & {\text{use the fundamental theorem of calculus part 1 with the chain rule }} \cr & {\text{to }}\frac{d}{{dx}}\left[ {\int_2^{{x^2}} {\sin \left( {{t^3}} \right)} dt} \right] \cr & \cr & \frac{{dy}}{{dx}} = x\left( {\sin {{\left( {{x^2}} \right)}^3}} \right)\frac{d}{{dx}}\left( {{x^2}} \right) + \int_2^{{x^2}} {\sin \left( {{t^3}} \right)} dt\frac{d}{{dx}}\left[ x \right] \cr & {\text{compute the derivatives and simplify}} \cr & \frac{{dy}}{{dx}} = x\left( {\sin {{\left( {{x^2}} \right)}^3}} \right)\left( {2x} \right) + \int_2^{{x^2}} {\sin \left( {{t^3}} \right)} dt\left( 1 \right) \cr & \frac{{dy}}{{dx}} = 2{x^2}\sin {x^6} + \int_2^{{x^2}} {\sin \left( {{t^3}} \right)} dt \cr} $$
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