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Derivative of exln⁡(x)\frac{e^{x}}{\ln\left(x\right)}

The derivative of exln⁡(x)\frac{e^{x}}{\ln\left(x\right)} is exln⁡(x)−ex1xln⁡(x)2\frac{e^{x} \ln\left(x\right) - e^{x} \frac{1}{x}}{\ln\left(x\right)^{2}}.

f(x) =
ddx[exln⁡(x)]=exln⁡(x)−ex1xln⁡(x)2\frac{d}{dx}\left[\frac{e^{x}}{\ln\left(x\right)}\right] = \frac{e^{x} \ln\left(x\right) - e^{x} \frac{1}{x}}{\ln\left(x\right)^{2}}

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Graph of f(x) = exp(x) / ln(x) and its derivative-3-2-1123-100-80-60-40-202040slope -9.242f(x)f'(x)tangent at x = 0.5
Graph of f(x) = exp(x) / ln(x) and its derivative. The dashed line is the tangent at x = 0.5; its slope is f'(0.5) = -9.2418.

How to work it out

  1. 1

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

    f(x)=exln⁡(x)f(x) = \frac{e^{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=exu = e^{x} and v=ln⁡(x)v = \ln\left(x\right).

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

    Differentiate each piece:

    u′=ex,v′=1xu' = e^{x},\quad v' = \frac{1}{x}
  4. 4

    Put them in the rule:

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

    Simplify:

    f′(x)=exln⁡(x)−ex1xln⁡(x)2f'(x) = \frac{e^{x} \ln\left(x\right) - e^{x} \frac{1}{x}}{\ln\left(x\right)^{2}}
  6. 6

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

    f′′(x)=(exln⁡(x)+ex1x−(ex1x+ex−1x2))ln⁡(x)2−(exln⁡(x)−ex1x)2ln⁡(x)1x(ln⁡(x)2)2f''(x) = \frac{\left(e^{x} \ln\left(x\right) + e^{x} \frac{1}{x} - \left(e^{x} \frac{1}{x} + e^{x} \frac{-1}{x^{2}}\right)\right) \ln\left(x\right)^{2} - \left(e^{x} \ln\left(x\right) - e^{x} \frac{1}{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
\[ \frac{d}{dx}\left[\frac{e^{x}}{\ln\left(x\right)}\right] = \frac{e^{x} \ln\left(x\right) - e^{x} \frac{1}{x}}{\ln\left(x\right)^{2}} \]
\[ f(x) = \frac{e^{x}}{\ln\left(x\right)} \]
\[ u = e^{x},\quad v = \ln\left(x\right) \]
\[ u' = e^{x},\quad v' = \frac{1}{x} \]
\[ f'(x) = \frac{\left(e^{x}\right)\left(\ln\left(x\right)\right) - \left(e^{x}\right)\left(\frac{1}{x}\right)}{\left(\ln\left(x\right)\right)^{2}} \]
\[ f'(x) = \frac{e^{x} \ln\left(x\right) - e^{x} \frac{1}{x}}{\ln\left(x\right)^{2}} \]
\[ f''(x) = \frac{\left(e^{x} \ln\left(x\right) + e^{x} \frac{1}{x} - \left(e^{x} \frac{1}{x} + e^{x} \frac{-1}{x^{2}}\right)\right) \ln\left(x\right)^{2} - \left(e^{x} \ln\left(x\right) - e^{x} \frac{1}{x}\right) 2 \ln\left(x\right) \frac{1}{x}}{\left(\ln\left(x\right)^{2}\right)^{2}} \]

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

xf(x)f'(x)
0.5-2.378602-9.2417966
210.6601552.9704785
318.28264412.73545

Questions

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

The derivative of exln⁡(x)\frac{e^{x}}{\ln\left(x\right)} is exln⁡(x)−ex1xln⁡(x)2\frac{e^{x} \ln\left(x\right) - e^{x} \frac{1}{x}}{\ln\left(x\right)^{2}}.

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

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

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

The quotient rule.

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

-9.2417966.

Related derivatives

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