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: 26


$$ y'(x) =\frac{(4 x-1) \cdot \sqrt{x^{4}+y^{4}}-2 x^{3}}{\sqrt{x^{4}+y^{4}}+2 y^{3}}$$

Work Step by Step

Given $$2x^2-x-y =\sqrt{ x^4+y^4}$$ Differentiate with respect to $x$ \begin{align*} \frac{d}{dx} \left(2x^2-x-y \right)&= \frac{d}{dx} \left(\sqrt{ x^4+y^4}\right)\\ 4 x-1-y^{\prime}&=\frac{1}{2 \sqrt{x^{4}+y^{4}}} \cdot\left(4 x^{3}+4 y^{3} \cdot y^{\prime}\right)\\ \left[-1-\frac{2 y^{3}}{\sqrt{x^{4}+y^{4}}}\right] y^{\prime}&=\frac{2 x^{3}}{\sqrt{x^{4}+y^{4}}}+1-4 x\\ \frac{\sqrt{x^{4}+y^{4}}+2 y^{3}}{\sqrt{x^{4}+y^{4}}} y^{\prime}&=\frac{(4 x-1) \cdot \sqrt{x^{4}+y^{4}}-2 x^{3}}{\sqrt{x^{4}+y^{4}}} \end{align*} Then $$ y'(x) =\frac{(4 x-1) \cdot \sqrt{x^{4}+y^{4}}-2 x^{3}}{\sqrt{x^{4}+y^{4}}+2 y^{3}}$$
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