Question

calculate the derivative of the following function.
y = 8(6x^4 + 5)^{-4}
$\frac{dy}{dx}=square$

Explanation:

Step1: Identify outer - inner functions

Let $u = 6x^{4}+5$, then $y = 8u^{-4}$.

Step2: Differentiate $y$ with respect to $u$

Using the power rule $\frac{d}{du}(au^{n})=nau^{n - 1}$, we have $\frac{dy}{du}=8\times(-4)u^{-5}=-32u^{-5}$.

Step3: Differentiate $u$ with respect to $x$

$\frac{du}{dx}=\frac{d}{dx}(6x^{4}+5)=24x^{3}$.

Step4: Apply the chain - rule

The chain - rule states that $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. Substitute $\frac{dy}{du}$ and $\frac{du}{dx}$: $\frac{dy}{dx}=-32u^{-5}\cdot24x^{3}$.

Step5: Substitute $u$ back

Replace $u = 6x^{4}+5$ into the above expression: $\frac{dy}{dx}=-32(6x^{4}+5)^{-5}\cdot24x^{3}=-768x^{3}(6x^{4}+5)^{-5}$.

Answer:

$-768x^{3}(6x^{4}+5)^{-5}$