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

The derivative of 1xln⁡(x)\frac{1}{x} \ln\left(x\right) is −1x2ln⁡(x)+1xx\frac{-1}{x^{2}} \ln\left(x\right) + \frac{\frac{1}{x}}{x}.

f(x) =
ddx[1xln⁡(x)]=−1x2ln⁡(x)+1xx\frac{d}{dx}\left[\frac{1}{x} \ln\left(x\right)\right] = \frac{-1}{x^{2}} \ln\left(x\right) + \frac{\frac{1}{x}}{x}

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Graph of f(x) = 1 / x * ln(x) and its derivative-3-2-1123-551015slope 1f(x)f'(x)tangent at x = 1
Graph of f(x) = 1 / x * ln(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)=1xln⁡(x)f(x) = \frac{1}{x} \ln\left(x\right) with respect to x. The rule that applies is the product rule.

    f(x)=1xln⁡(x)f(x) = \frac{1}{x} \ln\left(x\right)
  2. 2

    Product rule: (uv)′=u′v+uv′(uv)' = u'v + uv'. Let u=1xu = \frac{1}{x} and v=ln⁡(x)v = \ln\left(x\right).

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

    Differentiate each piece:

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

    Put them in the rule:

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

    Simplify:

    f′(x)=−1x2ln⁡(x)+1xxf'(x) = \frac{-1}{x^{2}} \ln\left(x\right) + \frac{\frac{1}{x}}{x}
  6. 6

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

    f′′(x)=−((−2)x)(x2)2ln⁡(x)+−1x21x+−1x2x−1xx2f''(x) = \frac{-\left(\left(-2\right) x\right)}{\left(x^{2}\right)^{2}} \ln\left(x\right) + \frac{-1}{x^{2}} \frac{1}{x} + \frac{\frac{-1}{x^{2}} x - \frac{1}{x}}{x^{2}}
LaTeX source of the formulas on this page
\[ \frac{d}{dx}\left[\frac{1}{x} \ln\left(x\right)\right] = \frac{-1}{x^{2}} \ln\left(x\right) + \frac{\frac{1}{x}}{x} \]
\[ f(x) = \frac{1}{x} \ln\left(x\right) \]
\[ u = \frac{1}{x},\quad v = \ln\left(x\right) \]
\[ u' = \frac{-1}{x^{2}},\quad v' = \frac{1}{x} \]
\[ f'(x) = \left(\frac{-1}{x^{2}}\right)\left(\ln\left(x\right)\right) + \left(\frac{1}{x}\right)\left(\frac{1}{x}\right) \]
\[ f'(x) = \frac{-1}{x^{2}} \ln\left(x\right) + \frac{\frac{1}{x}}{x} \]
\[ f''(x) = \frac{-\left(\left(-2\right) x\right)}{\left(x^{2}\right)^{2}} \ln\left(x\right) + \frac{-1}{x^{2}} \frac{1}{x} + \frac{\frac{-1}{x^{2}} x - \frac{1}{x}}{x^{2}} \]

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

xf(x)f'(x)
0.5-1.38629446.7725887
101
20.346573590.076713205
30.3662041-0.010956921

Questions

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

The derivative of 1xln⁡(x)\frac{1}{x} \ln\left(x\right) is −1x2ln⁡(x)+1xx\frac{-1}{x^{2}} \ln\left(x\right) + \frac{\frac{1}{x}}{x}.

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

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

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

The product rule.

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

1.

Related derivatives

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