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Derivative of ln⁡(x)1x\frac{\ln\left(x\right)}{\frac{1}{x}}

The derivative of ln⁡(x)1x\frac{\ln\left(x\right)}{\frac{1}{x}} is 1xx−ln⁡(x)−1x21x2\frac{\frac{\frac{1}{x}}{x} - \ln\left(x\right) \frac{-1}{x^{2}}}{\frac{1}{x}^{2}}.

f(x) =
ddx[ln⁡(x)1x]=1xx−ln⁡(x)−1x21x2\frac{d}{dx}\left[\frac{\ln\left(x\right)}{\frac{1}{x}}\right] = \frac{\frac{\frac{1}{x}}{x} - \ln\left(x\right) \frac{-1}{x^{2}}}{\frac{1}{x}^{2}}

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Graph of f(x) = ln(x) / (1 / x) and its derivative-3-2-1123-4-3-2-1123slope 1f(x)f'(x)tangent at x = 1
Graph of f(x) = ln(x) / (1 / x) 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)1xf(x) = \frac{\ln\left(x\right)}{\frac{1}{x}} with respect to x. The rule that applies is the quotient rule.

    f(x)=ln⁡(x)1xf(x) = \frac{\ln\left(x\right)}{\frac{1}{x}}
  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=1xv = \frac{1}{x}.

    u=ln⁡(x),v=1xu = \ln\left(x\right),\quad v = \frac{1}{x}
  3. 3

    Differentiate each piece:

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

    Put them in the rule:

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

    Simplify:

    f′(x)=1xx−ln⁡(x)−1x21x2f'(x) = \frac{\frac{\frac{1}{x}}{x} - \ln\left(x\right) \frac{-1}{x^{2}}}{\frac{1}{x}^{2}}
  6. 6

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

    f′′(x)=(−1x2x−1xx2−(−1x2x+ln⁡(x)−((−2)x)(x2)2))1x2−(1xx−ln⁡(x)−1x2)21x−1x2(1x2)2f''(x) = \frac{\left(\frac{\frac{-1}{x^{2}} x - \frac{1}{x}}{x^{2}} - \left(\frac{\frac{-1}{x^{2}}}{x} + \ln\left(x\right) \frac{-\left(\left(-2\right) x\right)}{\left(x^{2}\right)^{2}}\right)\right) \frac{1}{x}^{2} - \left(\frac{\frac{1}{x}}{x} - \ln\left(x\right) \frac{-1}{x^{2}}\right) 2 \frac{1}{x} \frac{-1}{x^{2}}}{\left(\frac{1}{x}^{2}\right)^{2}}
LaTeX source of the formulas on this page
\[ \frac{d}{dx}\left[\frac{\ln\left(x\right)}{\frac{1}{x}}\right] = \frac{\frac{\frac{1}{x}}{x} - \ln\left(x\right) \frac{-1}{x^{2}}}{\frac{1}{x}^{2}} \]
\[ f(x) = \frac{\ln\left(x\right)}{\frac{1}{x}} \]
\[ u = \ln\left(x\right),\quad v = \frac{1}{x} \]
\[ u' = \frac{1}{x},\quad v' = \frac{-1}{x^{2}} \]
\[ f'(x) = \frac{\left(\frac{1}{x}\right)\left(\frac{1}{x}\right) - \left(\ln\left(x\right)\right)\left(\frac{-1}{x^{2}}\right)}{\left(\frac{1}{x}\right)^{2}} \]
\[ f'(x) = \frac{\frac{\frac{1}{x}}{x} - \ln\left(x\right) \frac{-1}{x^{2}}}{\frac{1}{x}^{2}} \]
\[ f''(x) = \frac{\left(\frac{\frac{-1}{x^{2}} x - \frac{1}{x}}{x^{2}} - \left(\frac{\frac{-1}{x^{2}}}{x} + \ln\left(x\right) \frac{-\left(\left(-2\right) x\right)}{\left(x^{2}\right)^{2}}\right)\right) \frac{1}{x}^{2} - \left(\frac{\frac{1}{x}}{x} - \ln\left(x\right) \frac{-1}{x^{2}}\right) 2 \frac{1}{x} \frac{-1}{x^{2}}}{\left(\frac{1}{x}^{2}\right)^{2}} \]

Values of f and f' for ln(x) / (1 / x)

xf(x)f'(x)
0.5-0.346573590.30685282
101
21.38629441.6931472
33.29583692.0986123

Questions

What is the derivative of ln(x) / (1 / x)?

The derivative of ln⁡(x)1x\frac{\ln\left(x\right)}{\frac{1}{x}} is 1xx−ln⁡(x)−1x21x2\frac{\frac{\frac{1}{x}}{x} - \ln\left(x\right) \frac{-1}{x^{2}}}{\frac{1}{x}^{2}}.

What is the second derivative of ln(x) / (1 / x)?

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

Which rule do you use for ln(x) / (1 / x)?

The quotient rule.

What is the slope of ln(x) / (1 / x) at x = 1?

1.

Related derivatives

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