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

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

f(x) =
\frac{d}{dx}\left[\frac{\sqrt{x}}{e^{x}}\right] = \frac{\frac{e^{x}}{2 \sqrt{x}} - \sqrt{x} e^{x}}{e^{x}^{2}}

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Graph of f(x) = sqrt(x) / exp(x) and its derivative-3-2-1123-0.20.20.40.60.811.2slope -0.1839f(x)f'(x)tangent at x = 1
Graph of f(x) = sqrt(x) / exp(x) and its derivative. The dashed line is the tangent at x = 1; its slope is f'(1) = -0.18394.

How to work it out

  1. 1

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

    f(x)=xexf(x) = \frac{\sqrt{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=xu = \sqrt{x} and v=exv = e^{x}.

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

    Differentiate each piece:

    u′=12x,v′=exu' = \frac{1}{2 \sqrt{x}},\quad v' = e^{x}
  4. 4

    Put them in the rule:

    f′(x)=(12x)(ex)−(x)(ex)(ex)2f'(x) = \frac{\left(\frac{1}{2 \sqrt{x}}\right)\left(e^{x}\right) - \left(\sqrt{x}\right)\left(e^{x}\right)}{\left(e^{x}\right)^{2}}
  5. 5

    Simplify:

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

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

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

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

xf(x)f'(x)
0.50.42888194-8.1685645e-17
10.36787944-0.18393972
20.19139299-0.14354474
30.086233732-0.071861443

Questions

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

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

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

(((exp(x) * 2sqrt(x) - exp(x) * 2 * 1 / (2sqrt(x))) / (2sqrt(x))^2 - (exp(x) / (2sqrt(x)) + sqrt(x) * exp(x))) * exp(x)^2 - (exp(x) / (2sqrt(x)) - sqrt(x) * exp(x)) * 2exp(x) * exp(x)) / (exp(x)^2)^2.

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

The quotient rule.

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

-0.18393972.

Related derivatives

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