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 = $(-1-\frac{2x}{3})^{-3}$
u = -1 - $\frac{2x}{3}$ (type an expression using x as the variable.)
y = $2(-1-\frac{2x}{3})^{-4}$ (type an expression using u as the variable.)

Explanation:

Step1: Identify inner - outer functions

Let $u = - 1-\frac{2x}{3}$, then $y = u^{-3}$.

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

Using the power rule $\frac{d}{du}(u^n)=nu^{n - 1}$, for $y = u^{-3}$, we have $\frac{dy}{du}=-3u^{-4}$.

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

Differentiating $u=-1-\frac{2x}{3}$ with respect to $x$, we get $\frac{du}{dx}=-\frac{2}{3}$.

Step4: Use the chain - rule

The chain - rule states that $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. Substitute $\frac{dy}{du}=-3u^{-4}$ and $\frac{du}{dx}=-\frac{2}{3}$ into the chain - rule formula. Since $u=-1 - \frac{2x}{3}$, we have $\frac{dy}{dx}=(-3u^{-4})\cdot(-\frac{2}{3}) = 2u^{-4}=2(-1-\frac{2x}{3})^{-4}$.

Answer:

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