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Derivative of ln⁡(x)x5\frac{\ln\left(x\right)}{x^{5}}

The derivative of ln⁡(x)x5\frac{\ln\left(x\right)}{x^{5}} is x5x−ln⁡(x)5x4(x5)2\frac{\frac{x^{5}}{x} - \ln\left(x\right) 5 x^{4}}{\left(x^{5}\right)^{2}}.

f(x) =
ddx[ln⁡(x)x5]=x5x−ln⁡(x)5x4(x5)2\frac{d}{dx}\left[\frac{\ln\left(x\right)}{x^{5}}\right] = \frac{\frac{x^{5}}{x} - \ln\left(x\right) 5 x^{4}}{\left(x^{5}\right)^{2}}

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Graph of f(x) = ln(x) / x^5 and its derivative-3-2-1123-50050010001500200025003000slope 1f(x)f'(x)tangent at x = 1
Graph of f(x) = ln(x) / x^5 and its derivative. The dashed line is the tangent at x = 1; its slope is f'(1) = 1.

How to work it out

  1. 1

    We differentiate f(x)=ln⁡(x)x5f(x) = \frac{\ln\left(x\right)}{x^{5}} with respect to x. The rule that applies is the quotient rule.

    f(x)=ln⁡(x)x5f(x) = \frac{\ln\left(x\right)}{x^{5}}
  2. 2

    Quotient rule: (uv)′=u′v−uv′v2\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^{2}}. Let u=ln⁡(x)u = \ln\left(x\right) and v=x5v = x^{5}.

    u=ln⁡(x),v=x5u = \ln\left(x\right),\quad v = x^{5}
  3. 3

    Differentiate each piece:

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

    Put them in the rule:

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

    Simplify:

    f′(x)=x5x−ln⁡(x)5x4(x5)2f'(x) = \frac{\frac{x^{5}}{x} - \ln\left(x\right) 5 x^{4}}{\left(x^{5}\right)^{2}}
  6. 6

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

    f′′(x)=(5x4x−x5x2−(5x4x+ln⁡(x)20x3))(x5)2−(x5x−ln⁡(x)5x4)2x55x4((x5)2)2f''(x) = \frac{\left(\frac{5 x^{4} x - x^{5}}{x^{2}} - \left(\frac{5 x^{4}}{x} + \ln\left(x\right) 20 x^{3}\right)\right) \left(x^{5}\right)^{2} - \left(\frac{x^{5}}{x} - \ln\left(x\right) 5 x^{4}\right) 2 x^{5} 5 x^{4}}{\left(\left(x^{5}\right)^{2}\right)^{2}}
LaTeX source of the formulas on this page
\[ \frac{d}{dx}\left[\frac{\ln\left(x\right)}{x^{5}}\right] = \frac{\frac{x^{5}}{x} - \ln\left(x\right) 5 x^{4}}{\left(x^{5}\right)^{2}} \]
\[ f(x) = \frac{\ln\left(x\right)}{x^{5}} \]
\[ u = \ln\left(x\right),\quad v = x^{5} \]
\[ u' = \frac{1}{x},\quad v' = 5 x^{4} \]
\[ f'(x) = \frac{\left(\frac{1}{x}\right)\left(x^{5}\right) - \left(\ln\left(x\right)\right)\left(5 x^{4}\right)}{\left(x^{5}\right)^{2}} \]
\[ f'(x) = \frac{\frac{x^{5}}{x} - \ln\left(x\right) 5 x^{4}}{\left(x^{5}\right)^{2}} \]
\[ f''(x) = \frac{\left(\frac{5 x^{4} x - x^{5}}{x^{2}} - \left(\frac{5 x^{4}}{x} + \ln\left(x\right) 20 x^{3}\right)\right) \left(x^{5}\right)^{2} - \left(\frac{x^{5}}{x} - \ln\left(x\right) 5 x^{4}\right) 2 x^{5} 5 x^{4}}{\left(\left(x^{5}\right)^{2}\right)^{2}} \]

Values of f and f' for ln(x) / x5x^{5}

xf(x)f'(x)
0.5-22.18071285.8071
101
20.021660849-0.038527123
30.0045210382-0.0061633216

Questions

What is the derivative of ln(x) / x5x^{5}?

The derivative of ln⁡(x)x5\frac{\ln\left(x\right)}{x^{5}} is x5x−ln⁡(x)5x4(x5)2\frac{\frac{x^{5}}{x} - \ln\left(x\right) 5 x^{4}}{\left(x^{5}\right)^{2}}.

What is the second derivative of ln(x) / x5x^{5}?

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

Which rule do you use for ln(x) / x5x^{5}?

The quotient rule.

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

1.

Related derivatives

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