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Derivative of xx5\frac{\sqrt{x}}{x^{5}}

The derivative of xx5\frac{\sqrt{x}}{x^{5}} is x52x−x5x4(x5)2\frac{\frac{x^{5}}{2 \sqrt{x}} - \sqrt{x} 5 x^{4}}{\left(x^{5}\right)^{2}}.

f(x) =
ddx[xx5]=x52x−x5x4(x5)2\frac{d}{dx}\left[\frac{\sqrt{x}}{x^{5}}\right] = \frac{\frac{x^{5}}{2 \sqrt{x}} - \sqrt{x} 5 x^{4}}{\left(x^{5}\right)^{2}}

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Graph of f(x) = sqrt(x) / x^5 and its derivative-3-2-1123-1000-800-600-400-200200slope -4.5f(x)f'(x)tangent at x = 1
Graph of f(x) = sqrt(x) / x^5 and its derivative. The dashed line is the tangent at x = 1; its slope is f'(1) = -4.5.

How to work it out

  1. 1

    We differentiate f(x)=xx5f(x) = \frac{\sqrt{x}}{x^{5}} with respect to x. The rule that applies is the quotient rule.

    f(x)=xx5f(x) = \frac{\sqrt{x}}{x^{5}}
  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=x5v = x^{5}.

    u=x,v=x5u = \sqrt{x},\quad v = x^{5}
  3. 3

    Differentiate each piece:

    u′=12x,v′=5x4u' = \frac{1}{2 \sqrt{x}},\quad v' = 5 x^{4}
  4. 4

    Put them in the rule:

    f′(x)=(12x)(x5)−(x)(5x4)(x5)2f'(x) = \frac{\left(\frac{1}{2 \sqrt{x}}\right)\left(x^{5}\right) - \left(\sqrt{x}\right)\left(5 x^{4}\right)}{\left(x^{5}\right)^{2}}
  5. 5

    Simplify:

    f′(x)=x52x−x5x4(x5)2f'(x) = \frac{\frac{x^{5}}{2 \sqrt{x}} - \sqrt{x} 5 x^{4}}{\left(x^{5}\right)^{2}}
  6. 6

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

    f′′(x)=(5x42x−x5212x(2x)2−(5x42x+x20x3))(x5)2−(x52x−x5x4)2x55x4((x5)2)2f''(x) = \frac{\left(\frac{5 x^{4} 2 \sqrt{x} - x^{5} 2 \frac{1}{2 \sqrt{x}}}{\left(2 \sqrt{x}\right)^{2}} - \left(\frac{5 x^{4}}{2 \sqrt{x}} + \sqrt{x} 20 x^{3}\right)\right) \left(x^{5}\right)^{2} - \left(\frac{x^{5}}{2 \sqrt{x}} - \sqrt{x} 5 x^{4}\right) 2 x^{5} 5 x^{4}}{\left(\left(x^{5}\right)^{2}\right)^{2}}
LaTeX source of the formulas on this page
\[ \frac{d}{dx}\left[\frac{\sqrt{x}}{x^{5}}\right] = \frac{\frac{x^{5}}{2 \sqrt{x}} - \sqrt{x} 5 x^{4}}{\left(x^{5}\right)^{2}} \]
\[ f(x) = \frac{\sqrt{x}}{x^{5}} \]
\[ u = \sqrt{x},\quad v = x^{5} \]
\[ u' = \frac{1}{2 \sqrt{x}},\quad v' = 5 x^{4} \]
\[ f'(x) = \frac{\left(\frac{1}{2 \sqrt{x}}\right)\left(x^{5}\right) - \left(\sqrt{x}\right)\left(5 x^{4}\right)}{\left(x^{5}\right)^{2}} \]
\[ f'(x) = \frac{\frac{x^{5}}{2 \sqrt{x}} - \sqrt{x} 5 x^{4}}{\left(x^{5}\right)^{2}} \]
\[ f''(x) = \frac{\left(\frac{5 x^{4} 2 \sqrt{x} - x^{5} 2 \frac{1}{2 \sqrt{x}}}{\left(2 \sqrt{x}\right)^{2}} - \left(\frac{5 x^{4}}{2 \sqrt{x}} + \sqrt{x} 20 x^{3}\right)\right) \left(x^{5}\right)^{2} - \left(\frac{x^{5}}{2 \sqrt{x}} - \sqrt{x} 5 x^{4}\right) 2 x^{5} 5 x^{4}}{\left(\left(x^{5}\right)^{2}\right)^{2}} \]

Values of f and f' for sqrt(x) / x5x^{5}

xf(x)f'(x)
0.522.627417-203.64675
11-4.5
20.044194174-0.099436891
30.0071277811-0.010691672

Questions

What is the derivative of sqrt(x) / x5x^{5}?

The derivative of xx5\frac{\sqrt{x}}{x^{5}} is x52x−x5x4(x5)2\frac{\frac{x^{5}}{2 \sqrt{x}} - \sqrt{x} 5 x^{4}}{\left(x^{5}\right)^{2}}.

What is the second derivative of sqrt(x) / x5x^{5}?

(((5 * x4x^{4} * 2sqrt(x) - x5x^{5} * 2 * 1 / (2sqrt(x))) / (2sqrt(x))^2 - (5 * x4x^{4} / (2sqrt(x)) + sqrt(x) * 20 * x3x^{3})) * (x5x^{5})^2 - (x5x^{5} / (2sqrt(x)) - sqrt(x) * 5 * x4x^{4}) * 2 * x5x^{5} * 5 * x4x^{4}) / ((x5x^{5})^2)^2.

Which rule do you use for sqrt(x) / x5x^{5}?

The quotient rule.

What is the slope of sqrt(x) / x5x^{5} at x = 1?

-4.5.

Related derivatives

    Derivative of sqrt(x) / x^5: (x^5 / (2sqrt(x)) - sqrt(x) * 5 * x^4) / (x^5)^2 (with steps) | WordGPT