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Derivative of 1xx2\frac{\frac{1}{x}}{x^{2}}

The derivative of 1xx2\frac{\frac{1}{x}}{x^{2}} is −1x2x2−2xx(x2)2\frac{\frac{-1}{x^{2}} x^{2} - \frac{2 x}{x}}{\left(x^{2}\right)^{2}}.

f(x) =
ddx[1xx2]=−1x2x2−2xx(x2)2\frac{d}{dx}\left[\frac{\frac{1}{x}}{x^{2}}\right] = \frac{\frac{-1}{x^{2}} x^{2} - \frac{2 x}{x}}{\left(x^{2}\right)^{2}}

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Graph of f(x) = 1 / x / x^2 and its derivative-3-2-1123-1000-800-600-400-200slope -3f(x)f'(x)tangent at x = 1
Graph of f(x) = 1 / x / x^2 and its derivative. The dashed line is the tangent at x = 1; its slope is f'(1) = -3.

How to work it out

  1. 1

    We differentiate f(x)=1xx2f(x) = \frac{\frac{1}{x}}{x^{2}} with respect to x. The rule that applies is the quotient rule.

    f(x)=1xx2f(x) = \frac{\frac{1}{x}}{x^{2}}
  2. 2

    Quotient rule: (uv)′=u′v−uv′v2\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^{2}}. Let u=1xu = \frac{1}{x} and v=x2v = x^{2}.

    u=1x,v=x2u = \frac{1}{x},\quad v = x^{2}
  3. 3

    Differentiate each piece:

    u′=−1x2,v′=2xu' = \frac{-1}{x^{2}},\quad v' = 2 x
  4. 4

    Put them in the rule:

    f′(x)=(−1x2)(x2)−(1x)(2x)(x2)2f'(x) = \frac{\left(\frac{-1}{x^{2}}\right)\left(x^{2}\right) - \left(\frac{1}{x}\right)\left(2 x\right)}{\left(x^{2}\right)^{2}}
  5. 5

    Simplify:

    f′(x)=−1x2x2−2xx(x2)2f'(x) = \frac{\frac{-1}{x^{2}} x^{2} - \frac{2 x}{x}}{\left(x^{2}\right)^{2}}
  6. 6

    Differentiating once more gives the second derivative, which tells you about curvature:

    f′′(x)=(−((−2)x)(x2)2x2+−1x22x)(x2)2−(−1x2x2−2xx)2x22x((x2)2)2f''(x) = \frac{\left(\frac{-\left(\left(-2\right) x\right)}{\left(x^{2}\right)^{2}} x^{2} + \frac{-1}{x^{2}} 2 x\right) \left(x^{2}\right)^{2} - \left(\frac{-1}{x^{2}} x^{2} - \frac{2 x}{x}\right) 2 x^{2} 2 x}{\left(\left(x^{2}\right)^{2}\right)^{2}}
LaTeX source of the formulas on this page
\[ \frac{d}{dx}\left[\frac{\frac{1}{x}}{x^{2}}\right] = \frac{\frac{-1}{x^{2}} x^{2} - \frac{2 x}{x}}{\left(x^{2}\right)^{2}} \]
\[ f(x) = \frac{\frac{1}{x}}{x^{2}} \]
\[ u = \frac{1}{x},\quad v = x^{2} \]
\[ u' = \frac{-1}{x^{2}},\quad v' = 2 x \]
\[ f'(x) = \frac{\left(\frac{-1}{x^{2}}\right)\left(x^{2}\right) - \left(\frac{1}{x}\right)\left(2 x\right)}{\left(x^{2}\right)^{2}} \]
\[ f'(x) = \frac{\frac{-1}{x^{2}} x^{2} - \frac{2 x}{x}}{\left(x^{2}\right)^{2}} \]
\[ f''(x) = \frac{\left(\frac{-\left(\left(-2\right) x\right)}{\left(x^{2}\right)^{2}} x^{2} + \frac{-1}{x^{2}} 2 x\right) \left(x^{2}\right)^{2} - \left(\frac{-1}{x^{2}} x^{2} - \frac{2 x}{x}\right) 2 x^{2} 2 x}{\left(\left(x^{2}\right)^{2}\right)^{2}} \]

Values of f and f' for 1 / x / x2x^{2}

xf(x)f'(x)
-2-0.125-0.1875
-1-1-3
0.58-48
11-3
20.125-0.1875
30.037037037-0.037037037

Questions

What is the derivative of 1 / x / x2x^{2}?

The derivative of 1xx2\frac{\frac{1}{x}}{x^{2}} is −1x2x2−2xx(x2)2\frac{\frac{-1}{x^{2}} x^{2} - \frac{2 x}{x}}{\left(x^{2}\right)^{2}}.

What is the second derivative of 1 / x / x2x^{2}?

((-(-2x) / (x2x^{2})^2 * x2x^{2} + -1 / x2x^{2} * 2x) * (x2x^{2})^2 - (-1 / x2x^{2} * x2x^{2} - 2x / x) * 2 * x2x^{2} * 2x) / ((x2x^{2})^2)^2.

Which rule do you use for 1 / x / x2x^{2}?

The quotient rule.

What is the slope of 1 / x / x2x^{2} at x = 1?

-3.

Related derivatives

    Derivative of 1 / x / x^2: (-1 / x^2 * x^2 - 2x / x) / (x^2)^2 (with steps) | WordGPT