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: 16

Answer

$$f_x = -t^2e^{-x}.$$ $$f_t = 2te^{-x}.$$

Work Step by Step

To find $f_x$, we treat $t$ as a constant and differentiate with respect to $x$. Thus $$f_x = t^2 \cdot -e^{-x} = -t^2e^{-x}.$$ Similarly, to find $f_t$, we treat $x$ as constant and differentiate with respect to $t$. Thus $$f_t = 2te^{-x}.$$
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