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

The derivative of exx\frac{e^{x}}{\sqrt{x}} is exxāˆ’ex12xx2\frac{e^{x} \sqrt{x} - e^{x} \frac{1}{2 \sqrt{x}}}{\sqrt{x}^{2}}.

f(x) =
ddx[exx]=exxāˆ’ex12xx2\frac{d}{dx}\left[\frac{e^{x}}{\sqrt{x}}\right] = \frac{e^{x} \sqrt{x} - e^{x} \frac{1}{2 \sqrt{x}}}{\sqrt{x}^{2}}

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Graph of f(x) = exp(x) / sqrt(x) and its derivative-3-2-1123-4-2246810slope 1.359f(x)f'(x)tangent at x = 1
Graph of f(x) = exp(x) / sqrt(x) and its derivative. The dashed line is the tangent at x = 1; its slope is f'(1) = 1.3591.

How to work it out

  1. 1

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

    f(x)=exxf(x) = \frac{e^{x}}{\sqrt{x}}
  2. 2

    Quotient rule: (uv)ā€²=uā€²vāˆ’uvā€²v2\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^{2}}. Let u=exu = e^{x} and v=xv = \sqrt{x}.

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

    Differentiate each piece:

    uā€²=ex,vā€²=12xu' = e^{x},\quad v' = \frac{1}{2 \sqrt{x}}
  4. 4

    Put them in the rule:

    fā€²(x)=(ex)(x)āˆ’(ex)(12x)(x)2f'(x) = \frac{\left(e^{x}\right)\left(\sqrt{x}\right) - \left(e^{x}\right)\left(\frac{1}{2 \sqrt{x}}\right)}{\left(\sqrt{x}\right)^{2}}
  5. 5

    Simplify:

    fā€²(x)=exxāˆ’ex12xx2f'(x) = \frac{e^{x} \sqrt{x} - e^{x} \frac{1}{2 \sqrt{x}}}{\sqrt{x}^{2}}
  6. 6

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

    fā€²ā€²(x)=(exx+ex12xāˆ’(ex12x+exāˆ’(212x)(2x)2))x2āˆ’(exxāˆ’ex12x)2x12x(x2)2f''(x) = \frac{\left(e^{x} \sqrt{x} + e^{x} \frac{1}{2 \sqrt{x}} - \left(e^{x} \frac{1}{2 \sqrt{x}} + e^{x} \frac{-\left(2 \frac{1}{2 \sqrt{x}}\right)}{\left(2 \sqrt{x}\right)^{2}}\right)\right) \sqrt{x}^{2} - \left(e^{x} \sqrt{x} - e^{x} \frac{1}{2 \sqrt{x}}\right) 2 \sqrt{x} \frac{1}{2 \sqrt{x}}}{\left(\sqrt{x}^{2}\right)^{2}}
LaTeX source of the formulas on this page
\[ \frac{d}{dx}\left[\frac{e^{x}}{\sqrt{x}}\right] = \frac{e^{x} \sqrt{x} - e^{x} \frac{1}{2 \sqrt{x}}}{\sqrt{x}^{2}} \]
\[ f(x) = \frac{e^{x}}{\sqrt{x}} \]
\[ u = e^{x},\quad v = \sqrt{x} \]
\[ u' = e^{x},\quad v' = \frac{1}{2 \sqrt{x}} \]
\[ f'(x) = \frac{\left(e^{x}\right)\left(\sqrt{x}\right) - \left(e^{x}\right)\left(\frac{1}{2 \sqrt{x}}\right)}{\left(\sqrt{x}\right)^{2}} \]
\[ f'(x) = \frac{e^{x} \sqrt{x} - e^{x} \frac{1}{2 \sqrt{x}}}{\sqrt{x}^{2}} \]
\[ f''(x) = \frac{\left(e^{x} \sqrt{x} + e^{x} \frac{1}{2 \sqrt{x}} - \left(e^{x} \frac{1}{2 \sqrt{x}} + e^{x} \frac{-\left(2 \frac{1}{2 \sqrt{x}}\right)}{\left(2 \sqrt{x}\right)^{2}}\right)\right) \sqrt{x}^{2} - \left(e^{x} \sqrt{x} - e^{x} \frac{1}{2 \sqrt{x}}\right) 2 \sqrt{x} \frac{1}{2 \sqrt{x}}}{\left(\sqrt{x}^{2}\right)^{2}} \]

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

xf(x)f'(x)
0.52.3316444.4408921e-16
12.71828181.3591409
25.22485173.9186388
311.596399.6636585

Questions

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

The derivative of exx\frac{e^{x}}{\sqrt{x}} is exxāˆ’ex12xx2\frac{e^{x} \sqrt{x} - e^{x} \frac{1}{2 \sqrt{x}}}{\sqrt{x}^{2}}.

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

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

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

The quotient rule.

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

1.3591409.

Related derivatives

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