WordGPT

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

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

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

Type any function of x; the derivative is computed symbolically in your browser.

Explain it step by step in chatMake a short video of it
Graph of f(x) = 1 / x / ln(x) and its derivative-3-2-1123-30-20-1010slope -2.555f(x)f'(x)tangent at x = 0.5
Graph of f(x) = 1 / x / ln(x) and its derivative. The dashed line is the tangent at x = 0.5; its slope is f'(0.5) = -2.5547.

How to work it out

  1. 1

    We differentiate f(x)=1xln⁡(x)f(x) = \frac{\frac{1}{x}}{\ln\left(x\right)} with respect to x. The rule that applies is the quotient rule.

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

    Simplify:

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

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

    f′′(x)=(−((−2)x)(x2)2ln⁡(x)+−1x21x−−1x2x−1xx2)ln⁡(x)2−(−1x2ln⁡(x)−1xx)2ln⁡(x)1x(ln⁡(x)2)2f''(x) = \frac{\left(\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}}\right) \ln\left(x\right)^{2} - \left(\frac{-1}{x^{2}} \ln\left(x\right) - \frac{\frac{1}{x}}{x}\right) 2 \ln\left(x\right) \frac{1}{x}}{\left(\ln\left(x\right)^{2}\right)^{2}}
LaTeX source of the formulas on this page
\[ \frac{d}{dx}\left[\frac{\frac{1}{x}}{\ln\left(x\right)}\right] = \frac{\frac{-1}{x^{2}} \ln\left(x\right) - \frac{\frac{1}{x}}{x}}{\ln\left(x\right)^{2}} \]
\[ f(x) = \frac{\frac{1}{x}}{\ln\left(x\right)} \]
\[ u = \frac{1}{x},\quad v = \ln\left(x\right) \]
\[ u' = \frac{-1}{x^{2}},\quad v' = \frac{1}{x} \]
\[ f'(x) = \frac{\left(\frac{-1}{x^{2}}\right)\left(\ln\left(x\right)\right) - \left(\frac{1}{x}\right)\left(\frac{1}{x}\right)}{\left(\ln\left(x\right)\right)^{2}} \]
\[ f'(x) = \frac{\frac{-1}{x^{2}} \ln\left(x\right) - \frac{\frac{1}{x}}{x}}{\ln\left(x\right)^{2}} \]
\[ f''(x) = \frac{\left(\frac{-\left(\left(-2\right) x\right)}{\left(x^{2}\right)^{2}} \ln\left(x\right) + \frac{-1}{x^{2}} \frac{1}{x} - \frac{\frac{-1}{x^{2}} x - \frac{1}{x}}{x^{2}}\right) \ln\left(x\right)^{2} - \left(\frac{-1}{x^{2}} \ln\left(x\right) - \frac{\frac{1}{x}}{x}\right) 2 \ln\left(x\right) \frac{1}{x}}{\left(\ln\left(x\right)^{2}\right)^{2}} \]

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

xf(x)f'(x)
0.5-2.8853901-2.5546958
20.72134752-0.88101601
30.30341308-0.19319719

Questions

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

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

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

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

The quotient rule.

What is the slope of 1 / x / ln(x) at x = 0.5?

-2.5546958.

Related derivatives

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