suppose that\nfind $\frac{dy}{dx}$.\n$y = 5(x^{2}-2x)^{-4}$\n$\frac{dy}{dx}=$

Explanation:

Step1: Identify the outer - inner functions

Let $u = x^{2}-2x$, then $y = 5u^{-4}$.

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

Using the power rule $\frac{d}{du}(au^{n})=nau^{n - 1}$, we have $\frac{dy}{du}=5\times(-4)u^{-5}=-20u^{-5}$.

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

$\frac{du}{dx}=(x^{2}-2x)'=2x - 2$.

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 $\frac{du}{dx}$ into the chain - rule formula: $\frac{dy}{dx}=-20u^{-5}\cdot(2x - 2)$. Substitute $u = x^{2}-2x$ back into the expression: $\frac{dy}{dx}=-20(x^{2}-2x)^{-5}\cdot(2x - 2)=-20(2x - 2)(x^{2}-2x)^{-5}$.

Answer:

$-20(2x - 2)(x^{2}-2x)^{-5}$