Question

find the derivative of the following function.
y = \frac{-5}{(3x^{3}+5)^{2}}
\frac{dy}{dx}=\square

Explanation:

Step1: Rewrite the function

Rewrite $y = \frac{-5}{(3x^{3}+5)^{2}}$ as $y=-5(3x^{3}+5)^{-2}$.

Step2: Apply the chain - rule

The chain - rule states that if $y = f(g(x))$, then $y^\prime=f^\prime(g(x))\cdot g^\prime(x)$. Let $u = 3x^{3}+5$, so $y=-5u^{-2}$. First, find $\frac{dy}{du}$ and $\frac{du}{dx}$.
$\frac{dy}{du}=-5\times(-2)u^{-3}=10u^{-3}=\frac{10}{u^{3}}$, and $\frac{du}{dx}=9x^{2}$.

Step3: Calculate $\frac{dy}{dx}$

By the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. Substitute $u = 3x^{3}+5$ back in:
$\frac{dy}{dx}=\frac{10}{(3x^{3}+5)^{3}}\cdot9x^{2}=\frac{90x^{2}}{(3x^{3}+5)^{3}}$.

Answer:

$\frac{90x^{2}}{(3x^{3}+5)^{3}}$