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Derivative of x4x\frac{x^{4}}{\sqrt{x}}

The derivative of x4x\frac{x^{4}}{\sqrt{x}} is 4x3xāˆ’x412xx2\frac{4 x^{3} \sqrt{x} - x^{4} \frac{1}{2 \sqrt{x}}}{\sqrt{x}^{2}}.

f(x) =
ddx[x4x]=4x3xāˆ’x412xx2\frac{d}{dx}\left[\frac{x^{4}}{\sqrt{x}}\right] = \frac{4 x^{3} \sqrt{x} - x^{4} \frac{1}{2 \sqrt{x}}}{\sqrt{x}^{2}}

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Graph of f(x) = x^4 / sqrt(x) and its derivative-3-2-1123-101020304050slope 3.5f(x)f'(x)tangent at x = 1
Graph of f(x) = x^4 / sqrt(x) and its derivative. The dashed line is the tangent at x = 1; its slope is f'(1) = 3.5.

How to work it out

  1. 1

    We differentiate f(x)=x4xf(x) = \frac{x^{4}}{\sqrt{x}} with respect to x. The rule that applies is the quotient rule.

    f(x)=x4xf(x) = \frac{x^{4}}{\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=x4u = x^{4} and v=xv = \sqrt{x}.

    u=x4,v=xu = x^{4},\quad v = \sqrt{x}
  3. 3

    Differentiate each piece:

    uā€²=4x3,vā€²=12xu' = 4 x^{3},\quad v' = \frac{1}{2 \sqrt{x}}
  4. 4

    Put them in the rule:

    fā€²(x)=(4x3)(x)āˆ’(x4)(12x)(x)2f'(x) = \frac{\left(4 x^{3}\right)\left(\sqrt{x}\right) - \left(x^{4}\right)\left(\frac{1}{2 \sqrt{x}}\right)}{\left(\sqrt{x}\right)^{2}}
  5. 5

    Simplify:

    fā€²(x)=4x3xāˆ’x412xx2f'(x) = \frac{4 x^{3} \sqrt{x} - x^{4} \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)=(12x2x+4x312xāˆ’(4x312x+x4āˆ’(212x)(2x)2))x2āˆ’(4x3xāˆ’x412x)2x12x(x2)2f''(x) = \frac{\left(12 x^{2} \sqrt{x} + 4 x^{3} \frac{1}{2 \sqrt{x}} - \left(4 x^{3} \frac{1}{2 \sqrt{x}} + x^{4} \frac{-\left(2 \frac{1}{2 \sqrt{x}}\right)}{\left(2 \sqrt{x}\right)^{2}}\right)\right) \sqrt{x}^{2} - \left(4 x^{3} \sqrt{x} - x^{4} \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
\[ \frac{d}{dx}\left[\frac{x^{4}}{\sqrt{x}}\right] = \frac{4 x^{3} \sqrt{x} - x^{4} \frac{1}{2 \sqrt{x}}}{\sqrt{x}^{2}} \]
\[ f(x) = \frac{x^{4}}{\sqrt{x}} \]
\[ u = x^{4},\quad v = \sqrt{x} \]
\[ u' = 4 x^{3},\quad v' = \frac{1}{2 \sqrt{x}} \]
\[ f'(x) = \frac{\left(4 x^{3}\right)\left(\sqrt{x}\right) - \left(x^{4}\right)\left(\frac{1}{2 \sqrt{x}}\right)}{\left(\sqrt{x}\right)^{2}} \]
\[ f'(x) = \frac{4 x^{3} \sqrt{x} - x^{4} \frac{1}{2 \sqrt{x}}}{\sqrt{x}^{2}} \]
\[ f''(x) = \frac{\left(12 x^{2} \sqrt{x} + 4 x^{3} \frac{1}{2 \sqrt{x}} - \left(4 x^{3} \frac{1}{2 \sqrt{x}} + x^{4} \frac{-\left(2 \frac{1}{2 \sqrt{x}}\right)}{\left(2 \sqrt{x}\right)^{2}}\right)\right) \sqrt{x}^{2} - \left(4 x^{3} \sqrt{x} - x^{4} \frac{1}{2 \sqrt{x}}\right) 2 \sqrt{x} \frac{1}{2 \sqrt{x}}}{\left(\sqrt{x}^{2}\right)^{2}} \]

Values of f and f' for x4x^{4} / sqrt(x)

xf(x)f'(x)
0.50.0883883480.61871843
113.5
211.31370819.79899
346.76537254.5596

Questions

What is the derivative of x4x^{4} / sqrt(x)?

The derivative of x4x\frac{x^{4}}{\sqrt{x}} is 4x3xāˆ’x412xx2\frac{4 x^{3} \sqrt{x} - x^{4} \frac{1}{2 \sqrt{x}}}{\sqrt{x}^{2}}.

What is the second derivative of x4x^{4} / sqrt(x)?

((12 * x2x^{2} * sqrt(x) + 4 * x3x^{3} * 1 / (2sqrt(x)) - (4 * x3x^{3} * 1 / (2sqrt(x)) + x4x^{4} * -(2 * 1 / (2sqrt(x))) / (2sqrt(x))^2)) * sqrt(x)^2 - (4 * x3x^{3} * sqrt(x) - x4x^{4} * 1 / (2sqrt(x))) * 2sqrt(x) * 1 / (2sqrt(x))) / (sqrt(x)^2)^2.

Which rule do you use for x4x^{4} / sqrt(x)?

The quotient rule.

What is the slope of x4x^{4} / sqrt(x) at x = 1?

3.5.

Related derivatives

    Derivative of x^4 / sqrt(x): (4 * x^3 * sqrt(x) - x^4 * 1 / (2sqrt(x))) / sqrt(x)^2 (with steps) | WordGPT