Answer
a.
$f'(x) = -x^{-2}$
$f''(x) = 2x^{-3}$
$f'''(x) = -6 x^{-4}$
$f^{(n)}(x) = (-1)^n(n!)x^{-(n+1)}$
b. $f'(x) = -2x^{-3}$
$f''(x) = 6x^{-4}$
$f'''(x) = -24 x^{-5}$
$f^{(n)} (x) = (-1)^{n}(n+1)! x^{-(n+2)}$
Work Step by Step
a. We apply the power rule and notice a trend over time:
$f(x) = \frac{1}{x} = x^{-1}$
$f'(x) = -x^{-2}$
$f''(x) = 2x^{-3}$
$f'''(x) = -6 x^{-4}$
We notice that the coefficient of the $x$ term alternates between negative and positive and increases as a factorial. The exponent of $x$ is always negative and one less than the negative of the degree of the derivative (e.g. $f^{(3)}(x)$ has an exponent of $x^{-(3)-1} = x^{-4}$). Thus, $f^{(n)}(x) = (-1)^n(n!)x^{-(n+1)}$
b. We apply the power rule and notice a trend over time:
$f(x) = \frac{1}{x^2} = x^{-2}$
$f'(x) = -2x^{-3}$
$f''(x) = 6x^{-4}$
$f'''(x) = -24 x^{-5}$
We notice the coefficient of the $x$ term alternates between negative and positive and increases as a factorial (though it is one ahead of the previous example because we begin with $x^{-2}$ instead of $x^{-1}$. The exponent of $x$ is always negative and is two less than the negative of the degree of the derivative (e.g. $f^{(3)}(x)$ has an exponent of $x^{-(3)-2} = x^{-5}$). Thus, $f^{(n)} (x) = (-1)^{n}(n+1)! x^{-(n+2)}$
