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Derivative of 1xx5\frac{\frac{1}{x}}{x^{5}}

The derivative of 1xx5\frac{\frac{1}{x}}{x^{5}} is −1x2x5−5x4x(x5)2\frac{\frac{-1}{x^{2}} x^{5} - \frac{5 x^{4}}{x}}{\left(x^{5}\right)^{2}}.

f(x) =
ddx[1xx5]=−1x2x5−5x4x(x5)2\frac{d}{dx}\left[\frac{\frac{1}{x}}{x^{5}}\right] = \frac{\frac{-1}{x^{2}} x^{5} - \frac{5 x^{4}}{x}}{\left(x^{5}\right)^{2}}

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Graph of f(x) = 1 / x / x^5 and its derivative-3-2-1123200004000060000800001e+5slope -6f(x)f'(x)tangent at x = 1
Graph of f(x) = 1 / x / x^5 and its derivative. The dashed line is the tangent at x = 1; its slope is f'(1) = -6.

How to work it out

  1. 1

    We differentiate f(x)=1xx5f(x) = \frac{\frac{1}{x}}{x^{5}} with respect to x. The rule that applies is the quotient rule.

    f(x)=1xx5f(x) = \frac{\frac{1}{x}}{x^{5}}
  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=x5v = x^{5}.

    u=1x,v=x5u = \frac{1}{x},\quad v = x^{5}
  3. 3

    Differentiate each piece:

    u′=−1x2,v′=5x4u' = \frac{-1}{x^{2}},\quad v' = 5 x^{4}
  4. 4

    Put them in the rule:

    f′(x)=(−1x2)(x5)−(1x)(5x4)(x5)2f'(x) = \frac{\left(\frac{-1}{x^{2}}\right)\left(x^{5}\right) - \left(\frac{1}{x}\right)\left(5 x^{4}\right)}{\left(x^{5}\right)^{2}}
  5. 5

    Simplify:

    f′(x)=−1x2x5−5x4x(x5)2f'(x) = \frac{\frac{-1}{x^{2}} x^{5} - \frac{5 x^{4}}{x}}{\left(x^{5}\right)^{2}}
  6. 6

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

    f′′(x)=(−((−2)x)(x2)2x5+−1x25x4−20x3x−5x4x2)(x5)2−(−1x2x5−5x4x)2x55x4((x5)2)2f''(x) = \frac{\left(\frac{-\left(\left(-2\right) x\right)}{\left(x^{2}\right)^{2}} x^{5} + \frac{-1}{x^{2}} 5 x^{4} - \frac{20 x^{3} x - 5 x^{4}}{x^{2}}\right) \left(x^{5}\right)^{2} - \left(\frac{-1}{x^{2}} x^{5} - \frac{5 x^{4}}{x}\right) 2 x^{5} 5 x^{4}}{\left(\left(x^{5}\right)^{2}\right)^{2}}
LaTeX source of the formulas on this page
\[ \frac{d}{dx}\left[\frac{\frac{1}{x}}{x^{5}}\right] = \frac{\frac{-1}{x^{2}} x^{5} - \frac{5 x^{4}}{x}}{\left(x^{5}\right)^{2}} \]
\[ f(x) = \frac{\frac{1}{x}}{x^{5}} \]
\[ u = \frac{1}{x},\quad v = x^{5} \]
\[ u' = \frac{-1}{x^{2}},\quad v' = 5 x^{4} \]
\[ f'(x) = \frac{\left(\frac{-1}{x^{2}}\right)\left(x^{5}\right) - \left(\frac{1}{x}\right)\left(5 x^{4}\right)}{\left(x^{5}\right)^{2}} \]
\[ f'(x) = \frac{\frac{-1}{x^{2}} x^{5} - \frac{5 x^{4}}{x}}{\left(x^{5}\right)^{2}} \]
\[ f''(x) = \frac{\left(\frac{-\left(\left(-2\right) x\right)}{\left(x^{2}\right)^{2}} x^{5} + \frac{-1}{x^{2}} 5 x^{4} - \frac{20 x^{3} x - 5 x^{4}}{x^{2}}\right) \left(x^{5}\right)^{2} - \left(\frac{-1}{x^{2}} x^{5} - \frac{5 x^{4}}{x}\right) 2 x^{5} 5 x^{4}}{\left(\left(x^{5}\right)^{2}\right)^{2}} \]

Values of f and f' for 1 / x / x5x^{5}

xf(x)f'(x)
-20.0156250.046875
-116
0.564-768
11-6
20.015625-0.046875
30.0013717421-0.0027434842

Questions

What is the derivative of 1 / x / x5x^{5}?

The derivative of 1xx5\frac{\frac{1}{x}}{x^{5}} is −1x2x5−5x4x(x5)2\frac{\frac{-1}{x^{2}} x^{5} - \frac{5 x^{4}}{x}}{\left(x^{5}\right)^{2}}.

What is the second derivative of 1 / x / x5x^{5}?

((-(-2x) / (x2x^{2})^2 * x5x^{5} + -1 / x2x^{2} * 5 * x4x^{4} - (20 * x3x^{3} * x - 5 * x4x^{4}) / x2x^{2}) * (x5x^{5})^2 - (-1 / x2x^{2} * x5x^{5} - 5 * x4x^{4} / x) * 2 * x5x^{5} * 5 * x4x^{4}) / ((x5x^{5})^2)^2.

Which rule do you use for 1 / x / x5x^{5}?

The quotient rule.

What is the slope of 1 / x / x5x^{5} at x = 1?

-6.

Related derivatives

    Derivative of 1 / x / x^5: (-1 / x^2 * x^5 - 5 * x^4 / x) / (x^5)^2 (with steps) | WordGPT