#### 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 $z'$:
$4z^3z' = -4x^3$
$z^3z' = -x^3$
$z' = -\frac{x^3}{z^3}$

Published by
W. H. Freeman

ISBN 10:
1464125260

ISBN 13:
978-1-46412-526-3

$z' = -\frac{x^3}{z^3}$

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