Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 14 - Partial Derivatives - 14.3 Partial Derivatives - 14.3 Exercises: 42

Answer

The solution is $$f_y\left(1,\frac{1}{2}\right)=\frac{\pi+2\sqrt{3}}{6}.$$

Work Step by Step

First, find the derivative: $$f_y=\Big(y\sin^{-1}(xy))\Big)'_y=(y)'_y\sin^{-1}(xy)+y\big(\sin^{-1}(xy)\big)'_y=\sin^{-1}(xy)+y\frac{1}{\sqrt{1-(xy)^2}}\cdot(xy)'_y=\sin^{-1}(xy)+\frac{xy}{\sqrt{1-x^2y^2}}.$$ Now calculate by direct substitution $x=1$ and $y=1/2$: $$f_y\left(1,\frac{1}{2}\right)=\sin^{-1}\left(1\cdot\frac{1}{2}\right)+\frac{1\cdot\frac{1}{2}}{\sqrt{1-1^2\cdot\left(\frac{1}{2}\right)^2}}=\frac{\pi}{6}+\frac{\frac{1}{2}}{\sqrt{\frac{3}{4}}}=\frac{\pi}{6}+\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}=\frac{\pi+2\sqrt{3}}{6}.$$
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