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Derivative of x3x\frac{x^{3}}{\sqrt{x}}

The derivative of x3x\frac{x^{3}}{\sqrt{x}} is 3x2xāˆ’x312xx2\frac{3 x^{2} \sqrt{x} - x^{3} \frac{1}{2 \sqrt{x}}}{\sqrt{x}^{2}}.

f(x) =
ddx[x3x]=3x2xāˆ’x312xx2\frac{d}{dx}\left[\frac{x^{3}}{\sqrt{x}}\right] = \frac{3 x^{2} \sqrt{x} - x^{3} \frac{1}{2 \sqrt{x}}}{\sqrt{x}^{2}}

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Graph of f(x) = x^3 / sqrt(x) and its derivative-3-2-1123-10-551015slope 2.5f(x)f'(x)tangent at x = 1
Graph of f(x) = x^3 / sqrt(x) and its derivative. The dashed line is the tangent at x = 1; its slope is f'(1) = 2.5.

How to work it out

  1. 1

    We differentiate f(x)=x3xf(x) = \frac{x^{3}}{\sqrt{x}} with respect to x. The rule that applies is the quotient rule.

    f(x)=x3xf(x) = \frac{x^{3}}{\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=x3u = x^{3} and v=xv = \sqrt{x}.

    u=x3,v=xu = x^{3},\quad v = \sqrt{x}
  3. 3

    Differentiate each piece:

    uā€²=3x2,vā€²=12xu' = 3 x^{2},\quad v' = \frac{1}{2 \sqrt{x}}
  4. 4

    Put them in the rule:

    fā€²(x)=(3x2)(x)āˆ’(x3)(12x)(x)2f'(x) = \frac{\left(3 x^{2}\right)\left(\sqrt{x}\right) - \left(x^{3}\right)\left(\frac{1}{2 \sqrt{x}}\right)}{\left(\sqrt{x}\right)^{2}}
  5. 5

    Simplify:

    fā€²(x)=3x2xāˆ’x312xx2f'(x) = \frac{3 x^{2} \sqrt{x} - x^{3} \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)=(6xx+3x212xāˆ’(3x212x+x3āˆ’(212x)(2x)2))x2āˆ’(3x2xāˆ’x312x)2x12x(x2)2f''(x) = \frac{\left(6 x \sqrt{x} + 3 x^{2} \frac{1}{2 \sqrt{x}} - \left(3 x^{2} \frac{1}{2 \sqrt{x}} + x^{3} \frac{-\left(2 \frac{1}{2 \sqrt{x}}\right)}{\left(2 \sqrt{x}\right)^{2}}\right)\right) \sqrt{x}^{2} - \left(3 x^{2} \sqrt{x} - x^{3} \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
\[ \frac{d}{dx}\left[\frac{x^{3}}{\sqrt{x}}\right] = \frac{3 x^{2} \sqrt{x} - x^{3} \frac{1}{2 \sqrt{x}}}{\sqrt{x}^{2}} \]
\[ f(x) = \frac{x^{3}}{\sqrt{x}} \]
\[ u = x^{3},\quad v = \sqrt{x} \]
\[ u' = 3 x^{2},\quad v' = \frac{1}{2 \sqrt{x}} \]
\[ f'(x) = \frac{\left(3 x^{2}\right)\left(\sqrt{x}\right) - \left(x^{3}\right)\left(\frac{1}{2 \sqrt{x}}\right)}{\left(\sqrt{x}\right)^{2}} \]
\[ f'(x) = \frac{3 x^{2} \sqrt{x} - x^{3} \frac{1}{2 \sqrt{x}}}{\sqrt{x}^{2}} \]
\[ f''(x) = \frac{\left(6 x \sqrt{x} + 3 x^{2} \frac{1}{2 \sqrt{x}} - \left(3 x^{2} \frac{1}{2 \sqrt{x}} + x^{3} \frac{-\left(2 \frac{1}{2 \sqrt{x}}\right)}{\left(2 \sqrt{x}\right)^{2}}\right)\right) \sqrt{x}^{2} - \left(3 x^{2} \sqrt{x} - x^{3} \frac{1}{2 \sqrt{x}}\right) 2 \sqrt{x} \frac{1}{2 \sqrt{x}}}{\left(\sqrt{x}^{2}\right)^{2}} \]

Values of f and f' for x3x^{3} / sqrt(x)

xf(x)f'(x)
0.50.17677670.88388348
112.5
25.65685427.0710678
315.58845712.990381

Questions

What is the derivative of x3x^{3} / sqrt(x)?

The derivative of x3x\frac{x^{3}}{\sqrt{x}} is 3x2xāˆ’x312xx2\frac{3 x^{2} \sqrt{x} - x^{3} \frac{1}{2 \sqrt{x}}}{\sqrt{x}^{2}}.

What is the second derivative of x3x^{3} / sqrt(x)?

((6x * sqrt(x) + 3 * x2x^{2} * 1 / (2sqrt(x)) - (3 * x2x^{2} * 1 / (2sqrt(x)) + x3x^{3} * -(2 * 1 / (2sqrt(x))) / (2sqrt(x))^2)) * sqrt(x)^2 - (3 * x2x^{2} * sqrt(x) - x3x^{3} * 1 / (2sqrt(x))) * 2sqrt(x) * 1 / (2sqrt(x))) / (sqrt(x)^2)^2.

Which rule do you use for x3x^{3} / sqrt(x)?

The quotient rule.

What is the slope of x3x^{3} / sqrt(x) at x = 1?

2.5.

Related derivatives

    Derivative of x^3 / sqrt(x): (3 * x^2 * sqrt(x) - x^3 * 1 / (2sqrt(x))) / sqrt(x)^2 (with steps) | WordGPT