Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 15 - Differentiation in Several Variables - 15.6 The Chain Rule - Exercises - Page 809: 33

Answer

$$\nabla f =F^{\prime}(r) e_{\mathbf{r}}$$

Work Step by Step

Since \begin{aligned} \frac{\partial f}{\partial x}&=F^{\prime}(r) \frac{\partial r}{\partial x}\\ &=F^{\prime}(r) \cdot \frac{2 x}{2 \sqrt{x^{2}+y^{2}+z^{2}}}\\ &=F^{\prime}(r) \cdot \frac{x}{r}\\ \frac{\partial f}{\partial y}&=F^{\prime}(r) \frac{\partial r}{\partial y}\\ &=F^{\prime}(r) \cdot \frac{2 y}{2 \sqrt{x^{2}+y^{2}+z^{2}}}\\ &=F^{\prime}(r) \cdot \frac{y}{r}\\ \frac{\partial f}{\partial z}&=F^{\prime}(r) \frac{\partial r}{\partial z}\\ &=F^{\prime}(r) \cdot \frac{2 z}{2 \sqrt{x^{2}+y^{2}+z^{2}}}\\ &=F^{\prime}(r) \cdot \frac{z}{r} \end{aligned} Then \begin{align*} \nabla f&=\left\langle\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}\right\rangle\\ &=\left\langle F^{\prime}(r) \frac{x}{r}, F^{\prime}(r) \frac{y}{r}, F^{\prime}(r) \frac{z}{r}\right\rangle\\ &=\frac{F^{\prime}(r)}{r}\langle x, y, z\rangle\\ &= F^{\prime}(r) \frac{\mathbf{r}}{\|\mathbf{r}\|}\\ &=F^{\prime}(r) e_{\mathbf{r}} \end{align*}
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