Answer
a) $y' = \frac{10x}{3y^{2}}$
b) $y'=\frac{10x}{3(5x^2-7)^{2/3}}$
Work Step by Step
a) $5x^{2} - y^{3} = 7$
$10x - 3y^{2}(y') = 0$
$- 3y^{2}(y') = -10x$
$y' = \frac{-10x}{-3y^{2}}$
$y' = \frac{10x}{3y^{2}}$
b) $y^3=5x^2-7$
$y=\sqrt[3]{5x^2-7}$
$y'=\frac{10x}{3(5x^2-7)^{2/3}}$
c) We use (a) and (b):
$y'=\frac{10x}{3(\sqrt[3]{5x^2-7})^2}=\frac{10x}{3(5x^2-7)^{2/3}}$
