Single Variable Calculus, Chapter 3, 3.3, Section 3.3, Problem 60

Find the 1st and 2nd derivatives of $\displaystyle f(x) = \frac{1}{3-x}$


$
\begin{equation}
\begin{aligned}
f'(x) &= \frac{(3-x) \frac{d}{dx} (1) - \left[ (1) \frac{d}{dx}(3-x)\right]}{(3-x)^2} &&
\text{(Using Quotient Rule)}\\
f'(x) &= \frac{(3-x)(0)-(1)(-1)}{(3-x)^2} &&
\text{(Simplify the equation)}\\
\end{aligned}
\end{equation}
$


The first derivative of $f(x)$ is $\displaystyle f'(x) = \frac{1}{(3-x)^2}$


$
\begin{equation}
\begin{aligned}
f''(x) &= \frac{(3-x)^2 \frac{d}{dx}(1) - \left[ (1) \frac{d}{dx} (3-x)^2\right]}{[(3-x)^2]^2} &&
\text{(Using Quotient Rule)}\\
f''(x) &= \frac{(3-x)^2(0)-(1)(2)(-1)}{(3-x)^4} &&
\text{(Simplify the equation)}\\
\end{aligned}
\end{equation}
$


The second derivative of $f(x)$ is $\displaystyle f''(x) = \frac{2}{(3-x)^4}$