Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Review - Exercises: 53

Answer

$y''=-\dfrac{5x^{4}}{y^{11}}$

Work Step by Step

Find $y''$ if $x^{6}+y^{6}=1$ Use implicit differentiation to evaluate the first derivative: $x^{6}+y^{6}=1$ $6x^{5}+6y^{5}y'=0$ Solve for $y'$: $y'=-\dfrac{6x^{5}}{6y^{5}}$ $y'=-\dfrac{x^{5}}{y^{5}}$ Apply implicit differentiation again (and the quotient rule) to obtain the second derivative: $y''=-\dfrac{(y^{5})(x^{5})'-(x^{5})(y^{5})'}{y^{10}}=-\dfrac{5x^{4}y^{5}-5x^{5}y^{4}y'}{y^{10}}=...$ Substitute $y'$ by $-\dfrac{x^{5}}{y^{5}}$ and simplify: $...=-\dfrac{5x^{4}y^{5}-5x^{5}y^{4}\Big(-\dfrac{x^{5}}{y^{5}}\Big)}{y^{10}}=...$ $...=-\dfrac{5x^{4}y^{5}+\dfrac{5x^{10}}{y}}{y^{10}}=-\dfrac{\dfrac{5x^{4}y^{6}+5x^{10}}{y}}{y^{10}}=-\dfrac{5x^{4}y^{6}+5x^{10}}{y^{11}}=...$ $...=-\dfrac{5x^{4}(x^{6}+y^{6})}{y^{11}}=\displaystyle-\frac{5x^4}{y^{11}}$
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