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 $f''$