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Derivative of x1x\sqrt{x} \frac{1}{x}

The derivative of x1x\sqrt{x} \frac{1}{x} is 1x2x+xāˆ’1x2\frac{\frac{1}{x}}{2 \sqrt{x}} + \sqrt{x} \frac{-1}{x^{2}}.

f(x) =
ddx[x1x]=1x2x+xāˆ’1x2\frac{d}{dx}\left[\sqrt{x} \frac{1}{x}\right] = \frac{\frac{1}{x}}{2 \sqrt{x}} + \sqrt{x} \frac{-1}{x^{2}}

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Graph of f(x) = sqrt(x) * 1 / x and its derivative-3-2-1123-2-1123slope -0.5f(x)f'(x)tangent at x = 1
Graph of f(x) = sqrt(x) * 1 / x and its derivative. The dashed line is the tangent at x = 1; its slope is f'(1) = -0.5.

How to work it out

  1. 1

    We differentiate f(x)=x1xf(x) = \sqrt{x} \frac{1}{x} with respect to x. The rule that applies is the product rule.

    f(x)=x1xf(x) = \sqrt{x} \frac{1}{x}
  2. 2

    Product rule: (uv)ā€²=uā€²v+uvā€²(uv)' = u'v + uv'. Let u=xu = \sqrt{x} and v=1xv = \frac{1}{x}.

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

    Differentiate each piece:

    uā€²=12x,vā€²=āˆ’1x2u' = \frac{1}{2 \sqrt{x}},\quad v' = \frac{-1}{x^{2}}
  4. 4

    Put them in the rule:

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

    Simplify:

    fā€²(x)=1x2x+xāˆ’1x2f'(x) = \frac{\frac{1}{x}}{2 \sqrt{x}} + \sqrt{x} \frac{-1}{x^{2}}
  6. 6

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

    fā€²ā€²(x)=āˆ’1x22xāˆ’212xx(2x)2+āˆ’1x22x+xāˆ’((āˆ’2)x)(x2)2f''(x) = \frac{\frac{-1}{x^{2}} 2 \sqrt{x} - \frac{2 \frac{1}{2 \sqrt{x}}}{x}}{\left(2 \sqrt{x}\right)^{2}} + \frac{\frac{-1}{x^{2}}}{2 \sqrt{x}} + \sqrt{x} \frac{-\left(\left(-2\right) x\right)}{\left(x^{2}\right)^{2}}
LaTeX source of the formulas on this page
\[ \frac{d}{dx}\left[\sqrt{x} \frac{1}{x}\right] = \frac{\frac{1}{x}}{2 \sqrt{x}} + \sqrt{x} \frac{-1}{x^{2}} \]
\[ f(x) = \sqrt{x} \frac{1}{x} \]
\[ u = \sqrt{x},\quad v = \frac{1}{x} \]
\[ u' = \frac{1}{2 \sqrt{x}},\quad v' = \frac{-1}{x^{2}} \]
\[ f'(x) = \left(\frac{1}{2 \sqrt{x}}\right)\left(\frac{1}{x}\right) + \left(\sqrt{x}\right)\left(\frac{-1}{x^{2}}\right) \]
\[ f'(x) = \frac{\frac{1}{x}}{2 \sqrt{x}} + \sqrt{x} \frac{-1}{x^{2}} \]
\[ f''(x) = \frac{\frac{-1}{x^{2}} 2 \sqrt{x} - \frac{2 \frac{1}{2 \sqrt{x}}}{x}}{\left(2 \sqrt{x}\right)^{2}} + \frac{\frac{-1}{x^{2}}}{2 \sqrt{x}} + \sqrt{x} \frac{-\left(\left(-2\right) x\right)}{\left(x^{2}\right)^{2}} \]

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

xf(x)f'(x)
0.51.4142136-1.4142136
11-0.5
20.70710678-0.1767767
30.57735027-0.096225045

Questions

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

The derivative of x1x\sqrt{x} \frac{1}{x} is 1x2x+xāˆ’1x2\frac{\frac{1}{x}}{2 \sqrt{x}} + \sqrt{x} \frac{-1}{x^{2}}.

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

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

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

The product rule.

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

-0.5.

Related derivatives

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