Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 3 - Differentiation - 3.8 Implicit Differentiation - Exercises - Page 152: 22


$$ y'(x) = \frac{3 (x+y)^{2}-2xy\sec^2\left(x^{2} y\right) }{\left [x^{2}\sec^2 \left(x^{2} y\right) -3(x+y)^2 \right] } $$

Work Step by Step

Given $$\tan \left(x^{2} y\right)=(x+y)^{3} $$ Differentiate with respect to $x$ \begin{align*} \frac{d}{dx} \tan \left(x^{2} y\right)&=\frac{d}{dx} (x+y)^{3} \\ \sec^2 \left(x^{2} y\right)[x^{2} y'(x) +2x y ]&= 3 (x+y)^{2}( 1+y'(x) ) \\ \left [x^{2}\sec^2 \left(x^{2} y\right) -3(x+y)^2 \right] y'(x)& =3 (x+y)^{2}-2xy\sec^2 \left(x^{2} y\right) \end{align*} Then $$ y'(x) = \frac{3 (x+y)^{2}-2xy\sec^2\left(x^{2} y\right) }{\left [x^{2}\sec^2 \left(x^{2} y\right) -3(x+y)^2 \right] } $$
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