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Derivative of xx4\frac{\sqrt{x}}{x^{4}}

The derivative of xx4\frac{\sqrt{x}}{x^{4}} is x42x−x4x3(x4)2\frac{\frac{x^{4}}{2 \sqrt{x}} - \sqrt{x} 4 x^{3}}{\left(x^{4}\right)^{2}}.

f(x) =
ddx[xx4]=x42x−x4x3(x4)2\frac{d}{dx}\left[\frac{\sqrt{x}}{x^{4}}\right] = \frac{\frac{x^{4}}{2 \sqrt{x}} - \sqrt{x} 4 x^{3}}{\left(x^{4}\right)^{2}}

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Graph of f(x) = sqrt(x) / x^4 and its derivative-3-2-1123-300-250-200-150-100-5050slope -3.5f(x)f'(x)tangent at x = 1
Graph of f(x) = sqrt(x) / x^4 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)=xx4f(x) = \frac{\sqrt{x}}{x^{4}} with respect to x. The rule that applies is the quotient rule.

    f(x)=xx4f(x) = \frac{\sqrt{x}}{x^{4}}
  2. 2

    Quotient rule: (uv)′=u′v−uv′v2\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^{2}}. Let u=xu = \sqrt{x} and v=x4v = x^{4}.

    u=x,v=x4u = \sqrt{x},\quad v = x^{4}
  3. 3

    Differentiate each piece:

    u′=12x,v′=4x3u' = \frac{1}{2 \sqrt{x}},\quad v' = 4 x^{3}
  4. 4

    Put them in the rule:

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

    Simplify:

    f′(x)=x42x−x4x3(x4)2f'(x) = \frac{\frac{x^{4}}{2 \sqrt{x}} - \sqrt{x} 4 x^{3}}{\left(x^{4}\right)^{2}}
  6. 6

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

    f′′(x)=(4x32x−x4212x(2x)2−(4x32x+x12x2))(x4)2−(x42x−x4x3)2x44x3((x4)2)2f''(x) = \frac{\left(\frac{4 x^{3} 2 \sqrt{x} - x^{4} 2 \frac{1}{2 \sqrt{x}}}{\left(2 \sqrt{x}\right)^{2}} - \left(\frac{4 x^{3}}{2 \sqrt{x}} + \sqrt{x} 12 x^{2}\right)\right) \left(x^{4}\right)^{2} - \left(\frac{x^{4}}{2 \sqrt{x}} - \sqrt{x} 4 x^{3}\right) 2 x^{4} 4 x^{3}}{\left(\left(x^{4}\right)^{2}\right)^{2}}
LaTeX source of the formulas on this page
\[ \frac{d}{dx}\left[\frac{\sqrt{x}}{x^{4}}\right] = \frac{\frac{x^{4}}{2 \sqrt{x}} - \sqrt{x} 4 x^{3}}{\left(x^{4}\right)^{2}} \]
\[ f(x) = \frac{\sqrt{x}}{x^{4}} \]
\[ u = \sqrt{x},\quad v = x^{4} \]
\[ u' = \frac{1}{2 \sqrt{x}},\quad v' = 4 x^{3} \]
\[ f'(x) = \frac{\left(\frac{1}{2 \sqrt{x}}\right)\left(x^{4}\right) - \left(\sqrt{x}\right)\left(4 x^{3}\right)}{\left(x^{4}\right)^{2}} \]
\[ f'(x) = \frac{\frac{x^{4}}{2 \sqrt{x}} - \sqrt{x} 4 x^{3}}{\left(x^{4}\right)^{2}} \]
\[ f''(x) = \frac{\left(\frac{4 x^{3} 2 \sqrt{x} - x^{4} 2 \frac{1}{2 \sqrt{x}}}{\left(2 \sqrt{x}\right)^{2}} - \left(\frac{4 x^{3}}{2 \sqrt{x}} + \sqrt{x} 12 x^{2}\right)\right) \left(x^{4}\right)^{2} - \left(\frac{x^{4}}{2 \sqrt{x}} - \sqrt{x} 4 x^{3}\right) 2 x^{4} 4 x^{3}}{\left(\left(x^{4}\right)^{2}\right)^{2}} \]

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

xf(x)f'(x)
0.511.313708-79.195959
11-3.5
20.088388348-0.15467961
30.021383343-0.024947234

Questions

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

The derivative of xx4\frac{\sqrt{x}}{x^{4}} is x42x−x4x3(x4)2\frac{\frac{x^{4}}{2 \sqrt{x}} - \sqrt{x} 4 x^{3}}{\left(x^{4}\right)^{2}}.

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

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

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

The quotient rule.

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

-3.5.

Related derivatives

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