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 - Page 964: 18

Answer

$$f_x = \frac{3}{2\sqrt{3x + 4t}} .$$ $$f_t = \frac{2}{\sqrt{3x + 4t}} .$$

Work Step by Step

To find $f_x$, we treat $t$ as a constant and differentiate with respect to $x$. Thus $$f_x = \frac{1}{2}(3x + 4t)^{-1/2} \cdot \frac{\delta}{\delta x}(3x + 4t) = \frac{3}{\sqrt{3x + 4t}} .$$ Similarly, to find $f_t$, we treat $x$ as constant and differentiate with respect to $t$. Thus $$f_t = \frac{1}{2}(3x + 4t)^{-1/2} \frac{\delta}{\delta t}(3x + 4t) = \frac{4}{2\sqrt{3x + 4t}} = \frac{2}{\sqrt{3x + 4t}}.$$
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