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Derivative of 1xx\frac{\frac{1}{x}}{\sqrt{x}}

The derivative of 1xx\frac{\frac{1}{x}}{\sqrt{x}} is āˆ’1x2xāˆ’12xxx2\frac{\frac{-1}{x^{2}} \sqrt{x} - \frac{\frac{1}{2 \sqrt{x}}}{x}}{\sqrt{x}^{2}}.

f(x) =
ddx[1xx]=āˆ’1x2xāˆ’12xxx2\frac{d}{dx}\left[\frac{\frac{1}{x}}{\sqrt{x}}\right] = \frac{\frac{-1}{x^{2}} \sqrt{x} - \frac{\frac{1}{2 \sqrt{x}}}{x}}{\sqrt{x}^{2}}

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Graph of f(x) = 1 / x / sqrt(x) and its derivative-3-2-1123-20-15-10-5510slope -1.5f(x)f'(x)tangent at x = 1
Graph of f(x) = 1 / x / sqrt(x) and its derivative. The dashed line is the tangent at x = 1; its slope is f'(1) = -1.5.

How to work it out

  1. 1

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

    f(x)=1xxf(x) = \frac{\frac{1}{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=1xu = \frac{1}{x} and v=xv = \sqrt{x}.

    u=1x,v=xu = \frac{1}{x},\quad v = \sqrt{x}
  3. 3

    Differentiate each piece:

    uā€²=āˆ’1x2,vā€²=12xu' = \frac{-1}{x^{2}},\quad v' = \frac{1}{2 \sqrt{x}}
  4. 4

    Put them in the rule:

    fā€²(x)=(āˆ’1x2)(x)āˆ’(1x)(12x)(x)2f'(x) = \frac{\left(\frac{-1}{x^{2}}\right)\left(\sqrt{x}\right) - \left(\frac{1}{x}\right)\left(\frac{1}{2 \sqrt{x}}\right)}{\left(\sqrt{x}\right)^{2}}
  5. 5

    Simplify:

    fā€²(x)=āˆ’1x2xāˆ’12xxx2f'(x) = \frac{\frac{-1}{x^{2}} \sqrt{x} - \frac{\frac{1}{2 \sqrt{x}}}{x}}{\sqrt{x}^{2}}
  6. 6

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

    fā€²ā€²(x)=(āˆ’((āˆ’2)x)(x2)2x+āˆ’1x212xāˆ’āˆ’(212x)(2x)2xāˆ’12xx2)x2āˆ’(āˆ’1x2xāˆ’12xx)2x12x(x2)2f''(x) = \frac{\left(\frac{-\left(\left(-2\right) x\right)}{\left(x^{2}\right)^{2}} \sqrt{x} + \frac{-1}{x^{2}} \frac{1}{2 \sqrt{x}} - \frac{\frac{-\left(2 \frac{1}{2 \sqrt{x}}\right)}{\left(2 \sqrt{x}\right)^{2}} x - \frac{1}{2 \sqrt{x}}}{x^{2}}\right) \sqrt{x}^{2} - \left(\frac{-1}{x^{2}} \sqrt{x} - \frac{\frac{1}{2 \sqrt{x}}}{x}\right) 2 \sqrt{x} \frac{1}{2 \sqrt{x}}}{\left(\sqrt{x}^{2}\right)^{2}}
LaTeX source of the formulas on this page
\[ \frac{d}{dx}\left[\frac{\frac{1}{x}}{\sqrt{x}}\right] = \frac{\frac{-1}{x^{2}} \sqrt{x} - \frac{\frac{1}{2 \sqrt{x}}}{x}}{\sqrt{x}^{2}} \]
\[ f(x) = \frac{\frac{1}{x}}{\sqrt{x}} \]
\[ u = \frac{1}{x},\quad v = \sqrt{x} \]
\[ u' = \frac{-1}{x^{2}},\quad v' = \frac{1}{2 \sqrt{x}} \]
\[ f'(x) = \frac{\left(\frac{-1}{x^{2}}\right)\left(\sqrt{x}\right) - \left(\frac{1}{x}\right)\left(\frac{1}{2 \sqrt{x}}\right)}{\left(\sqrt{x}\right)^{2}} \]
\[ f'(x) = \frac{\frac{-1}{x^{2}} \sqrt{x} - \frac{\frac{1}{2 \sqrt{x}}}{x}}{\sqrt{x}^{2}} \]
\[ f''(x) = \frac{\left(\frac{-\left(\left(-2\right) x\right)}{\left(x^{2}\right)^{2}} \sqrt{x} + \frac{-1}{x^{2}} \frac{1}{2 \sqrt{x}} - \frac{\frac{-\left(2 \frac{1}{2 \sqrt{x}}\right)}{\left(2 \sqrt{x}\right)^{2}} x - \frac{1}{2 \sqrt{x}}}{x^{2}}\right) \sqrt{x}^{2} - \left(\frac{-1}{x^{2}} \sqrt{x} - \frac{\frac{1}{2 \sqrt{x}}}{x}\right) 2 \sqrt{x} \frac{1}{2 \sqrt{x}}}{\left(\sqrt{x}^{2}\right)^{2}} \]

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

xf(x)f'(x)
0.52.8284271-8.4852814
11-1.5
20.35355339-0.26516504
30.19245009-0.096225045

Questions

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

The derivative of 1xx\frac{\frac{1}{x}}{\sqrt{x}} is āˆ’1x2xāˆ’12xxx2\frac{\frac{-1}{x^{2}} \sqrt{x} - \frac{\frac{1}{2 \sqrt{x}}}{x}}{\sqrt{x}^{2}}.

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

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

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

The quotient rule.

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

-1.5.

Related derivatives

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