Calculus: Early Transcendentals 8th Edition

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

Chapter 14 - Section 14.3 - Partial Derivatives - 14.3 Exercise: 66

Answer

$g_{rst}=e^{r}cos(st)-ste^{r}sin(st)$

Work Step by Step

Consider the function $g(r,s,t)=e^{r} sin(st)$ Let us start by finding $g_{r}(r,s,t)$ by differentiating $g(r,s,t) $ with respect to $r$ keeping $s$ and $t$ constant. As we know $g_{r}=\frac{∂}{∂r}g(r,s,t) $ $=\frac{∂}{∂r}[e^{r} sin(st)]$ $=e^{r} sin(st)$ Differentiate $g_{r}(r,s,t)$ with respect to $s$ keeping $r$ and $t$ constant . $g_{rs}=\frac{∂}{∂s}[e^{r} sin(st)]=e^{r}tcos(st)$ Now, $g_{rst}=\frac{∂}{∂t}[e^{r}tcos(st)]=-e^{r}stsin(st)+e^{r}cos(st)$ Hence, $g_{rst}=e^{r}cos(st)-ste^{r}sin(st)$
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