Answer
$z' = -\frac{x^3}{z^3}$
Work Step by Step
$\frac{d}{dx}(x^4) + \frac{d}{dx}(z^3) = \frac{d}{dx}(1)$
$4x^3 + 4z^3z' = 0$
Now we just need to solve for $z'$:
$4z^3z' = -4x^3$
$z^3z' = -x^3$
$z' = -\frac{x^3}{z^3}$
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
Published by
W. H. Freeman
ISBN 10:
1464125260
ISBN 13:
978-1-46412-526-3
Chapter 3 - Differentiation - 3.8 Implicit Differentiation - Exercises - Page 152: 14
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