Answer
$y'= \frac{-315x^2}{(7x^3+1)^4}$
Work Step by Step
$\frac{d}{dx}[f(g(x))] = f'(g(x)) \times g'(x)$
$y=5(7x^3+1)^{-3}$
Because 5 is a constant, the answer will be $5 \times \frac{d}{dx}[(7x^3+1)^{-3}]$
$\frac{d}{dx}[(7x^3+1)^{-3}] = (-3)(7x^3+1)^{-4} \times 21x^2$
$5 \times \frac{d}{dx}[(7x^3+1)^{-3}] = (5) (-3)(7x^3+1)^{-4} \times 21x^2 = \frac{-315x^2}{(7x^3+1)^4}$
