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: 43

Answer

The solution is $$f_y(1,2,2)=\frac{1}{6}.$$

Work Step by Step

First, find the derivative: $$f_y=\left(\ln\frac{1-\sqrt{x^2+y^2+z^2}}{1+\sqrt{x^2+y^2+z^2}}\right)'_y=\frac{1}{\frac{1-\sqrt{x^2+y^2+z^2}}{1+\sqrt{x^2+y^2+z^2}}}=\left(\frac{1-\sqrt{x^2+y^2+z^2}}{1+\sqrt{x^2+y^2+z^2}}\right)'_y=\frac{1+\sqrt{x^2+y^2+z^2}}{1-\sqrt{x^2+y^2+z^2}}\times\\ \times\frac{(1-\sqrt{x^2+y^2+z^2})'_y(1+\sqrt{x^2+y^2+z^2})-(1-\sqrt{x^2+y^2+z^2})(1+\sqrt{x^2+y^2+z^2})'_y}{(1+\sqrt{x^2+y^2+z^2})^2}=\\ \frac{1}{1-\sqrt{x^2+y^2+z^2}}\times\\ \times\frac{(1-\sqrt{x^2+y^2+z^2})'_y(1+\sqrt{x^2+y^2+z^2})-(1-\sqrt{x^2+y^2+z^2})(1+\sqrt{x^2+y^2+z^2})'_y}{(1+\sqrt{x^2+y^2+z^2})}.$$ Now we will separately calculate $(\sqrt{x^2+y^2+z^2})'_y=\frac{1}{2\sqrt{x^2+y^2+z^2}}(x^2+y^2+z^2)'_y=\frac{2y}{2\sqrt{x^2+y^2+z^2}}=\frac{y}{\sqrt{x^2+y^2+z^2}};$ then also: $(1-\sqrt{x^2+y^2+z^2})'_y=-\frac{y}{\sqrt{x^2+y^2+z^2}},$ and $(1+\sqrt{x^2+y^2+z^2})'_y=\frac{y}{\sqrt{x^2+y^2+z^2}}.$ This further gives: $(1-\sqrt{x^2+y^2+z^2})'_y(1+\sqrt{x^2+y^2+z^2})=-\frac{y}{\sqrt{x^2+y^2+z^2}}-y$ and $(1+\sqrt{x^2+y^2+z^2})'_y(1-\sqrt{x^2+y^2+z^2})=\frac{y}{\sqrt{x^2+y^2+z^2}}-y$. Putting this together: $$(1-\sqrt{x^2+y^2+z^2})'_y(1+\sqrt{x^2+y^2+z^2})-(1+\sqrt{x^2+y^2+z^2})'_y(1-\sqrt{x^2+y^2+z^2})=\\ -\frac{y}{\sqrt{x^2+y^2+z^2}}-y-\left(\frac{y}{\sqrt{x^2+y^2+z^2}}-y\right)=\\ -\frac{2y}{\sqrt{x^2+y^2+z^2}}.$$ Putting all of this into the derivatve we get: $$f_y=\frac{1}{1-\sqrt{x^2+y^2+z^2}}\cdot\frac{\frac{-2y}{\sqrt{x^2+y^2+z^2}}}{1+\sqrt{x^2+y^2+z^2}}=\frac{2y}{(x^2+y^2+z^2-1)\sqrt{x^2+y^2+z^2}}.$$ Finally, lets calculate by direct substitution $x=1$, $y=2$, $z=2$: $$f_y(1,2,2)=\frac{2\cdot2}{(1^2+2^2+2^2-1)\sqrt{1^2+2^2+2^2}}=\frac{4}{8\cdot3}=\frac{1}{6}.$$
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