d/dx (1/(3 - x)) equals -ln|3 - x| -1/(3 - x)^2 1/(3 - x)^2 ln|3 - x|

Answer:

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

Explanation:

Step1: Rewrite the function

Let $y=\frac{1}{3 - x}=(3 - x)^{-1}$.

Step2: Apply chain - rule

The chain - rule states that if $y = f(g(x))$, then $y^\prime=f^\prime(g(x))\cdot g^\prime(x)$. Let $u = 3 - x$, so $y = u^{-1}$. First, find $\frac{dy}{du}$ and $\frac{du}{dx}$. $\frac{dy}{du}=-1\times u^{-2}=-\frac{1}{u^{2}}$ and $\frac{du}{dx}=-1$.

Step3: Calculate the derivative

By the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. Substitute $u = 3 - x$ back in: $\frac{dy}{dx}=-\frac{1}{u^{2}}\times(-1)=\frac{1}{(3 - x)^{2}}$.