Question

find the derivative of the following function.
y = \frac{-5}{(4x^{5}+9)^{3}}
\frac{dy}{dx}=\square

Explanation:

Step1: Rewrite the function

Rewrite $y = \frac{- 5}{(4x^{5}+9)^{3}}$ as $y=-5(4x^{5}+9)^{-3}$.

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 $u = 4x^{5}+9$, so $y=-5u^{-3}$. First, find $\frac{dy}{du}$ and $\frac{du}{dx}$.
$\frac{dy}{du}=-5\times(-3)u^{-4}=15u^{-4}$.
$\frac{du}{dx}=20x^{4}$.

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

By the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. Substitute $u = 4x^{5}+9$ back in:
$\frac{dy}{dx}=15(4x^{5}+9)^{-4}\times20x^{4}=\frac{300x^{4}}{(4x^{5}+9)^{4}}$.

Answer:

$\frac{300x^{4}}{(4x^{5}+9)^{4}}$