Question

question
find $\frac{dy}{dx}$ if $y = (-2x^{3}-4)^{-4}$.
provide your answer below:
$\frac{dy}{dx}=square$

Explanation:

Step1: Identify the outer - function and inner - function

Let $u=-2x^{3}-4$, then $y = u^{-4}$.

Step2: Differentiate the outer - function with respect to $u$

Using the power rule $\frac{d}{du}(u^{n})=nu^{n - 1}$, we have $\frac{dy}{du}=-4u^{-5}$.

Step3: Differentiate the inner - function with respect to $x$

$\frac{du}{dx}=-6x^{2}$.

Step4: Apply the chain rule

The chain rule states that $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. Substituting $\frac{dy}{du}$ and $\frac{du}{dx}$ into the chain - rule formula, we get $\frac{dy}{dx}=-4u^{-5}\cdot(-6x^{2})$.

Step5: Substitute $u=-2x^{3}-4$ back into the expression

$\frac{dy}{dx}=-4(-2x^{3}-4)^{-5}\cdot(-6x^{2})=\frac{24x^{2}}{(-2x^{3}-4)^{5}}$.

Answer:

$\frac{24x^{2}}{(-2x^{3}-4)^{5}}$