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Derivative of 1xx3\frac{\frac{1}{x}}{x^{3}}

The derivative of 1xx3\frac{\frac{1}{x}}{x^{3}} is −1x2x3−3x2x(x3)2\frac{\frac{-1}{x^{2}} x^{3} - \frac{3 x^{2}}{x}}{\left(x^{3}\right)^{2}}.

f(x) =
ddx[1xx3]=−1x2x3−3x2x(x3)2\frac{d}{dx}\left[\frac{\frac{1}{x}}{x^{3}}\right] = \frac{\frac{-1}{x^{2}} x^{3} - \frac{3 x^{2}}{x}}{\left(x^{3}\right)^{2}}

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Graph of f(x) = 1 / x / x^3 and its derivative-3-2-112350010001500200025003000slope -4f(x)f'(x)tangent at x = 1
Graph of f(x) = 1 / x / x^3 and its derivative. The dashed line is the tangent at x = 1; its slope is f'(1) = -4.

How to work it out

  1. 1

    We differentiate f(x)=1xx3f(x) = \frac{\frac{1}{x}}{x^{3}} with respect to x. The rule that applies is the quotient rule.

    f(x)=1xx3f(x) = \frac{\frac{1}{x}}{x^{3}}
  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=x3v = x^{3}.

    u=1x,v=x3u = \frac{1}{x},\quad v = x^{3}
  3. 3

    Differentiate each piece:

    u′=−1x2,v′=3x2u' = \frac{-1}{x^{2}},\quad v' = 3 x^{2}
  4. 4

    Put them in the rule:

    f′(x)=(−1x2)(x3)−(1x)(3x2)(x3)2f'(x) = \frac{\left(\frac{-1}{x^{2}}\right)\left(x^{3}\right) - \left(\frac{1}{x}\right)\left(3 x^{2}\right)}{\left(x^{3}\right)^{2}}
  5. 5

    Simplify:

    f′(x)=−1x2x3−3x2x(x3)2f'(x) = \frac{\frac{-1}{x^{2}} x^{3} - \frac{3 x^{2}}{x}}{\left(x^{3}\right)^{2}}
  6. 6

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

    f′′(x)=(−((−2)x)(x2)2x3+−1x23x2−6xx−3x2x2)(x3)2−(−1x2x3−3x2x)2x33x2((x3)2)2f''(x) = \frac{\left(\frac{-\left(\left(-2\right) x\right)}{\left(x^{2}\right)^{2}} x^{3} + \frac{-1}{x^{2}} 3 x^{2} - \frac{6 x x - 3 x^{2}}{x^{2}}\right) \left(x^{3}\right)^{2} - \left(\frac{-1}{x^{2}} x^{3} - \frac{3 x^{2}}{x}\right) 2 x^{3} 3 x^{2}}{\left(\left(x^{3}\right)^{2}\right)^{2}}
LaTeX source of the formulas on this page
\[ \frac{d}{dx}\left[\frac{\frac{1}{x}}{x^{3}}\right] = \frac{\frac{-1}{x^{2}} x^{3} - \frac{3 x^{2}}{x}}{\left(x^{3}\right)^{2}} \]
\[ f(x) = \frac{\frac{1}{x}}{x^{3}} \]
\[ u = \frac{1}{x},\quad v = x^{3} \]
\[ u' = \frac{-1}{x^{2}},\quad v' = 3 x^{2} \]
\[ f'(x) = \frac{\left(\frac{-1}{x^{2}}\right)\left(x^{3}\right) - \left(\frac{1}{x}\right)\left(3 x^{2}\right)}{\left(x^{3}\right)^{2}} \]
\[ f'(x) = \frac{\frac{-1}{x^{2}} x^{3} - \frac{3 x^{2}}{x}}{\left(x^{3}\right)^{2}} \]
\[ f''(x) = \frac{\left(\frac{-\left(\left(-2\right) x\right)}{\left(x^{2}\right)^{2}} x^{3} + \frac{-1}{x^{2}} 3 x^{2} - \frac{6 x x - 3 x^{2}}{x^{2}}\right) \left(x^{3}\right)^{2} - \left(\frac{-1}{x^{2}} x^{3} - \frac{3 x^{2}}{x}\right) 2 x^{3} 3 x^{2}}{\left(\left(x^{3}\right)^{2}\right)^{2}} \]

Values of f and f' for 1 / x / x3x^{3}

xf(x)f'(x)
-20.06250.125
-114
0.516-128
11-4
20.0625-0.125
30.012345679-0.016460905

Questions

What is the derivative of 1 / x / x3x^{3}?

The derivative of 1xx3\frac{\frac{1}{x}}{x^{3}} is −1x2x3−3x2x(x3)2\frac{\frac{-1}{x^{2}} x^{3} - \frac{3 x^{2}}{x}}{\left(x^{3}\right)^{2}}.

What is the second derivative of 1 / x / x3x^{3}?

((-(-2x) / (x2x^{2})^2 * x3x^{3} + -1 / x2x^{2} * 3 * x2x^{2} - (6x * x - 3 * x2x^{2}) / x2x^{2}) * (x3x^{3})^2 - (-1 / x2x^{2} * x3x^{3} - 3 * x2x^{2} / x) * 2 * x3x^{3} * 3 * x2x^{2}) / ((x3x^{3})^2)^2.

Which rule do you use for 1 / x / x3x^{3}?

The quotient rule.

What is the slope of 1 / x / x3x^{3} at x = 1?

-4.

Related derivatives

    Derivative of 1 / x / x^3: (-1 / x^2 * x^3 - 3 * x^2 / x) / (x^3)^2 (with steps) | WordGPT