Question

write the function in the form y = f(u) and u = g(x). then find $\frac{dy}{dx}$ as a function of x.
$y=left(-1 - \frac{2x}{3}
ight)^{-3}$
$u=-1 - \frac{2x}{3}$ (type an expression using x as the variable.)
$y = 2left(-1-\frac{2x}{3}
ight)^{-4}$ (type an expression using u as the variable.)

Explanation:

Step1: Identify u and y in terms of u

Given $u = - 1-\frac{2x}{3}$ and $y = u^{-3}$.

Step2: Find $\frac{du}{dx}$

Differentiate $u=-1 - \frac{2x}{3}$ with respect to $x$. Using the power - rule for differentiation, $\frac{du}{dx}=-\frac{2}{3}$.

Step3: Find $\frac{dy}{du}$

Differentiate $y = u^{-3}$ with respect to $u$. By the power - rule $\frac{dy}{du}=-3u^{-4}$.

Step4: Use the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$

Substitute $\frac{dy}{du}=-3u^{-4}$ and $\frac{du}{dx}=-\frac{2}{3}$ into the chain - rule formula. Then $\frac{dy}{dx}=(-3u^{-4})\cdot(-\frac{2}{3})$.
Replace $u=-1-\frac{2x}{3}$ back into the expression for $\frac{dy}{dx}$. So $\frac{dy}{dx}=2(-1 - \frac{2x}{3})^{-4}$.

Answer:

$\frac{dy}{dx}=2(-1-\frac{2x}{3})^{-4}$