WordGPT

Derivative of 1xex\frac{\frac{1}{x}}{e^{x}}

The derivative of 1xex\frac{\frac{1}{x}}{e^{x}} is \frac{\frac{-1}{x^{2}} e^{x} - \frac{e^{x}}{x}}{e^{x}^{2}}.

f(x) =
\frac{d}{dx}\left[\frac{\frac{1}{x}}{e^{x}}\right] = \frac{\frac{-1}{x^{2}} e^{x} - \frac{e^{x}}{x}}{e^{x}^{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 / exp(x) and its derivative-3-2-1123-15-10-55slope -0.7358f(x)f'(x)tangent at x = 1
Graph of f(x) = 1 / x / exp(x) and its derivative. The dashed line is the tangent at x = 1; its slope is f'(1) = -0.73576.

How to work it out

  1. 1

    We differentiate f(x)=1xexf(x) = \frac{\frac{1}{x}}{e^{x}} with respect to x. The rule that applies is the quotient rule.

    f(x)=1xexf(x) = \frac{\frac{1}{x}}{e^{x}}
  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=exv = e^{x}.

    u=1x,v=exu = \frac{1}{x},\quad v = e^{x}
  3. 3

    Differentiate each piece:

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

    Put them in the rule:

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

    Simplify:

    f'(x) = \frac{\frac{-1}{x^{2}} e^{x} - \frac{e^{x}}{x}}{e^{x}^{2}}
  6. 6

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

    f''(x) = \frac{\left(\frac{-\left(\left(-2\right) x\right)}{\left(x^{2}\right)^{2}} e^{x} + \frac{-1}{x^{2}} e^{x} - \frac{e^{x} x - e^{x}}{x^{2}}\right) e^{x}^{2} - \left(\frac{-1}{x^{2}} e^{x} - \frac{e^{x}}{x}\right) 2 e^{x} e^{x}}{\left(e^{x}^{2}\right)^{2}}
LaTeX source of the formulas on this page
\[ \frac{d}{dx}\left[\frac{\frac{1}{x}}{e^{x}}\right] = \frac{\frac{-1}{x^{2}} e^{x} - \frac{e^{x}}{x}}{e^{x}^{2}} \]
\[ f(x) = \frac{\frac{1}{x}}{e^{x}} \]
\[ u = \frac{1}{x},\quad v = e^{x} \]
\[ u' = \frac{-1}{x^{2}},\quad v' = e^{x} \]
\[ f'(x) = \frac{\left(\frac{-1}{x^{2}}\right)\left(e^{x}\right) - \left(\frac{1}{x}\right)\left(e^{x}\right)}{\left(e^{x}\right)^{2}} \]
\[ f'(x) = \frac{\frac{-1}{x^{2}} e^{x} - \frac{e^{x}}{x}}{e^{x}^{2}} \]
\[ f''(x) = \frac{\left(\frac{-\left(\left(-2\right) x\right)}{\left(x^{2}\right)^{2}} e^{x} + \frac{-1}{x^{2}} e^{x} - \frac{e^{x} x - e^{x}}{x^{2}}\right) e^{x}^{2} - \left(\frac{-1}{x^{2}} e^{x} - \frac{e^{x}}{x}\right) 2 e^{x} e^{x}}{\left(e^{x}^{2}\right)^{2}} \]

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

xf(x)f'(x)
-2-3.6945281.847264
-1-2.71828180
0.51.2130613-3.639184
10.36787944-0.73575888
20.067667642-0.10150146
30.016595689-0.022127586

Questions

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

The derivative of 1xex\frac{\frac{1}{x}}{e^{x}} is \frac{\frac{-1}{x^{2}} e^{x} - \frac{e^{x}}{x}}{e^{x}^{2}}.

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

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

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

The quotient rule.

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

-0.73575888.

Related derivatives

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