Single Variable Calculus, Chapter 3, 3.2, Section 3.2, Problem 44

Find $f'(x)$ and $f''(x)$ on the function $f(x) = \displaystyle \frac{1}{x}$ using the definition of a derivative. Then graph $f, f'$ and $f''$ on a common screen and check to see if your answers are reasonable.

Using the definition of derivative

Using the definition of derivative


$\begin{equation} \begin{aligned} \qquad f'(x) =& \lim_{h \to 0} \frac{f(x + h) = f(x)}{h} && \\ \\ \qquad f'(x) =& \lim_{h \to 0} \frac{\displaystyle \frac{1}{x + h} - \frac{1}{x}}{h} && \text{Substitute $f(x + h)$ and $f(x)$} \\ \\ \qquad f'(x) =& \lim_{h \to 0} \frac{x - (x + h)}{(h)(x) (x + h)} && \text{Get the LCD on the numerator and simplify} \\ \\ \qquad f'(x) =& \lim_{h \to 0} \frac{\cancel{x} - \cancel{x} - h}{(h)(x)(x + h)} && \text{Combine like terms} \\ \\ \qquad f'(x) =& \lim_{h \to 0} \frac{-\cancel{h}}{\cancel{(h)} (x)(x + h)} && \text{Cancel out like terms} \\ \\ f'(x) =& \lim_{h \to 0} \left[ \frac{-1}{(x)(x + h)} \right] = \frac{-1}{(x)(x + 0)} = \frac{-1}{(x)(x)} && \text{Evaluate the limit} \\ \\ f'(x) =& \frac{-1}{x^2} && \end{aligned} \end{equation}$


Using the 2nd derivative of the definition


$\begin{equation} \begin{aligned} \qquad f''(x) =& \lim_{h \to 0} \frac{f'(x + h) = f'(x)}{h} && \\ \\ \qquad f''(x) =& \lim_{h \to 0} \frac{\displaystyle \frac{-1}{(x + h)^2} - \left( \frac{-1}{x^2} \right)}{h} && \text{Substitute } f'(x + h) \text{ and } f'(x) \\ \\ \qquad f''(x) =& \lim_{h \to 0} \frac{-x^2 + (x + h)^2}{(h)(x^2)(x + h)^2} && \text{Get the LCD on the numerator and simplify} \\ \\ \qquad f''(x) =& \lim_{h \to 0} \frac{-x^2 + x^2 + 2xh + h^2}{(h)(x^2)(x + h)^2} && \text{Expand the equation} \\ \\ \qquad f''(x) =& \lim_{h \to 0} \frac{\cancel{-x^2} + \cancel{x^2} + 2xh + h^2}{(h)(x^2)(x + h)^2} && \text{Combine like terms} \\ \\ \qquad f''(x) =& \lim_{h \to 0} \frac{2xh + h^2}{(h)(x^2)(x + h)^2} && \text{Factor the numerator} \\ \\ \qquad f''(x) =& \lim_{h \to 0} \frac{\cancel{h}(2x + h)}{\cancel{(h)}(x^2)(x + h)^2} && \text{Cancel out like terms} \\ \\ \qquad f''(x) =& \lim_{h \to 0} \left[ \frac{2x + h}{(x^2)(x + h)^2} \right] = \frac{2x + 0}{(x^2)(x + 0)^2} = \frac{2x}{(x^2)(x^2)} && \text{Evaluate the limit} \\ \\ \qquad f''(x) =& \frac{2x}{x^4} \end{aligned} \end{equation}$


Graph $f, f'$ and $f''$