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 $\displaystyle f''(x) = \frac{2}{(3-x)^4}$