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Derivative of x2x\frac{x^{2}}{\sqrt{x}}

The derivative of x2x\frac{x^{2}}{\sqrt{x}} is 2xxāˆ’x212xx2\frac{2 x \sqrt{x} - x^{2} \frac{1}{2 \sqrt{x}}}{\sqrt{x}^{2}}.

f(x) =
ddx[x2x]=2xxāˆ’x212xx2\frac{d}{dx}\left[\frac{x^{2}}{\sqrt{x}}\right] = \frac{2 x \sqrt{x} - x^{2} \frac{1}{2 \sqrt{x}}}{\sqrt{x}^{2}}

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Graph of f(x) = x^2 / sqrt(x) and its derivative-3-2-1123-4-224slope 1.5f(x)f'(x)tangent at x = 1
Graph of f(x) = x^2 / 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)=x2xf(x) = \frac{x^{2}}{\sqrt{x}} with respect to x. The rule that applies is the quotient rule.

    f(x)=x2xf(x) = \frac{x^{2}}{\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=x2u = x^{2} and v=xv = \sqrt{x}.

    u=x2,v=xu = x^{2},\quad v = \sqrt{x}
  3. 3

    Differentiate each piece:

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

    Put them in the rule:

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

    Simplify:

    fā€²(x)=2xxāˆ’x212xx2f'(x) = \frac{2 x \sqrt{x} - x^{2} \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)=(2x+2x12xāˆ’(2x12x+x2āˆ’(212x)(2x)2))x2āˆ’(2xxāˆ’x212x)2x12x(x2)2f''(x) = \frac{\left(2 \sqrt{x} + 2 x \frac{1}{2 \sqrt{x}} - \left(2 x \frac{1}{2 \sqrt{x}} + x^{2} \frac{-\left(2 \frac{1}{2 \sqrt{x}}\right)}{\left(2 \sqrt{x}\right)^{2}}\right)\right) \sqrt{x}^{2} - \left(2 x \sqrt{x} - x^{2} \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
\[ \frac{d}{dx}\left[\frac{x^{2}}{\sqrt{x}}\right] = \frac{2 x \sqrt{x} - x^{2} \frac{1}{2 \sqrt{x}}}{\sqrt{x}^{2}} \]
\[ f(x) = \frac{x^{2}}{\sqrt{x}} \]
\[ u = x^{2},\quad v = \sqrt{x} \]
\[ u' = 2 x,\quad v' = \frac{1}{2 \sqrt{x}} \]
\[ f'(x) = \frac{\left(2 x\right)\left(\sqrt{x}\right) - \left(x^{2}\right)\left(\frac{1}{2 \sqrt{x}}\right)}{\left(\sqrt{x}\right)^{2}} \]
\[ f'(x) = \frac{2 x \sqrt{x} - x^{2} \frac{1}{2 \sqrt{x}}}{\sqrt{x}^{2}} \]
\[ f''(x) = \frac{\left(2 \sqrt{x} + 2 x \frac{1}{2 \sqrt{x}} - \left(2 x \frac{1}{2 \sqrt{x}} + x^{2} \frac{-\left(2 \frac{1}{2 \sqrt{x}}\right)}{\left(2 \sqrt{x}\right)^{2}}\right)\right) \sqrt{x}^{2} - \left(2 x \sqrt{x} - x^{2} \frac{1}{2 \sqrt{x}}\right) 2 \sqrt{x} \frac{1}{2 \sqrt{x}}}{\left(\sqrt{x}^{2}\right)^{2}} \]

Values of f and f' for x2x^{2} / sqrt(x)

xf(x)f'(x)
0.50.353553391.0606602
111.5
22.82842712.1213203
35.19615242.5980762

Questions

What is the derivative of x2x^{2} / sqrt(x)?

The derivative of x2x\frac{x^{2}}{\sqrt{x}} is 2xxāˆ’x212xx2\frac{2 x \sqrt{x} - x^{2} \frac{1}{2 \sqrt{x}}}{\sqrt{x}^{2}}.

What is the second derivative of x2x^{2} / sqrt(x)?

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

Which rule do you use for x2x^{2} / sqrt(x)?

The quotient rule.

What is the slope of x2x^{2} / sqrt(x) at x = 1?

1.5.

Related derivatives

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