Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 13 - Partial Derivatives - 13.5 The Chain Rule - Exercises Set 13.5 - Page 958: 54

Answer

$\left(\frac{\partial w}{\partial \rho}\right)^{2}=\left(\frac{\partial w}{\partial z}\right)^{2}+\left(\frac{\partial w}{\partial y}\right)^{2}+\left(\frac{\partial w}{\partial x}\right)^{2}$

Work Step by Step

Let $f$ be a differential function of one variable \[ \left(x^{2}+y^{2}+z^{2}\right)^{1 / 2}=\rho \] Let $(\rho)f=w$ \[ \begin{array}{l} \frac{\partial w}{\partial x}=\frac{d w}{d \rho} \frac{\partial \rho}{\partial x} \\ \frac{\partial w}{\partial y}=\frac{d w}{d \rho} \frac{\partial \rho}{\partial y} \\ \frac{\partial w}{\partial z}=\frac{d w}{d \rho} \frac{\partial \rho}{\partial z} \end{array} \] Let's consider $\frac{\partial \rho}{\partial x}=\frac{2 x}{2\left(x^{2}+y^{2}+z^{2}\right)^{1 / 2}}, \quad \frac{\partial \rho}{\partial y}=\frac{2 y}{2\left(x^{2}+y^{2}+z^{2}\right)^{1 / 2}}$ and $\quad \frac{\partial \rho}{\partial z}=\frac{2 z}{2\left(x^{2}+y^{2}+z^{2}\right)^{1 / 2}}$ Thus, here we can write \[ \left(\frac{\partial w}{\partial x}\right)^{2}+\left(\frac{\partial w}{\partial y}\right)^{2}+\left(\frac{\partial w}{\partial z}\right)^{2}=\left(\frac{\partial w}{\partial \rho}\right)^{2} \frac{x^{2}}{\left(x^{2}+y^{2}+z^{2}\right)}+\left(\frac{\partial w}{\partial \rho}\right)^{2} \frac{y^{2}}{\left(x^{2}+y^{2}+z^{2}\right)}+\left(\frac{\partial w}{\partial \rho}\right)^{2} \frac{z^{2}}{\left(x^{2}+y^{2}+z^{2}\right)} \] $\left(\frac{\partial w}{\partial x}\right)^{2}+\left(\frac{\partial w}{\partial y}\right)^{2}+\left(\frac{\partial w}{\partial z}\right)^{2}=\left(\frac{\partial w}{\partial \rho}\right)^{2}\left(\frac{x^{2}+y^{2}+z^{2}}{x^{2}+y^{2}+z^{2}}\right)$ $\left(\frac{\partial w}{\partial \rho}\right)^{2}=\left(\frac{\partial w}{\partial z}\right)^{2}+\left(\frac{\partial w}{\partial y}\right)^{2}+\left(\frac{\partial w}{\partial x}\right)^{2}$
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