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Derivative of xx3\frac{\sqrt{x}}{x^{3}}

The derivative of xx3\frac{\sqrt{x}}{x^{3}} is x32x−x3x2(x3)2\frac{\frac{x^{3}}{2 \sqrt{x}} - \sqrt{x} 3 x^{2}}{\left(x^{3}\right)^{2}}.

f(x) =
ddx[xx3]=x32x−x3x2(x3)2\frac{d}{dx}\left[\frac{\sqrt{x}}{x^{3}}\right] = \frac{\frac{x^{3}}{2 \sqrt{x}} - \sqrt{x} 3 x^{2}}{\left(x^{3}\right)^{2}}

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Graph of f(x) = sqrt(x) / x^3 and its derivative-3-2-1123-80-60-40-2020slope -2.5f(x)f'(x)tangent at x = 1
Graph of f(x) = sqrt(x) / x^3 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)=xx3f(x) = \frac{\sqrt{x}}{x^{3}} with respect to x. The rule that applies is the quotient rule.

    f(x)=xx3f(x) = \frac{\sqrt{x}}{x^{3}}
  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=x3v = x^{3}.

    u=x,v=x3u = \sqrt{x},\quad v = x^{3}
  3. 3

    Differentiate each piece:

    u′=12x,v′=3x2u' = \frac{1}{2 \sqrt{x}},\quad v' = 3 x^{2}
  4. 4

    Put them in the rule:

    f′(x)=(12x)(x3)−(x)(3x2)(x3)2f'(x) = \frac{\left(\frac{1}{2 \sqrt{x}}\right)\left(x^{3}\right) - \left(\sqrt{x}\right)\left(3 x^{2}\right)}{\left(x^{3}\right)^{2}}
  5. 5

    Simplify:

    f′(x)=x32x−x3x2(x3)2f'(x) = \frac{\frac{x^{3}}{2 \sqrt{x}} - \sqrt{x} 3 x^{2}}{\left(x^{3}\right)^{2}}
  6. 6

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

    f′′(x)=(3x22x−x3212x(2x)2−(3x22x+x6x))(x3)2−(x32x−x3x2)2x33x2((x3)2)2f''(x) = \frac{\left(\frac{3 x^{2} 2 \sqrt{x} - x^{3} 2 \frac{1}{2 \sqrt{x}}}{\left(2 \sqrt{x}\right)^{2}} - \left(\frac{3 x^{2}}{2 \sqrt{x}} + \sqrt{x} 6 x\right)\right) \left(x^{3}\right)^{2} - \left(\frac{x^{3}}{2 \sqrt{x}} - \sqrt{x} 3 x^{2}\right) 2 x^{3} 3 x^{2}}{\left(\left(x^{3}\right)^{2}\right)^{2}}
LaTeX source of the formulas on this page
\[ \frac{d}{dx}\left[\frac{\sqrt{x}}{x^{3}}\right] = \frac{\frac{x^{3}}{2 \sqrt{x}} - \sqrt{x} 3 x^{2}}{\left(x^{3}\right)^{2}} \]
\[ f(x) = \frac{\sqrt{x}}{x^{3}} \]
\[ u = \sqrt{x},\quad v = x^{3} \]
\[ u' = \frac{1}{2 \sqrt{x}},\quad v' = 3 x^{2} \]
\[ f'(x) = \frac{\left(\frac{1}{2 \sqrt{x}}\right)\left(x^{3}\right) - \left(\sqrt{x}\right)\left(3 x^{2}\right)}{\left(x^{3}\right)^{2}} \]
\[ f'(x) = \frac{\frac{x^{3}}{2 \sqrt{x}} - \sqrt{x} 3 x^{2}}{\left(x^{3}\right)^{2}} \]
\[ f''(x) = \frac{\left(\frac{3 x^{2} 2 \sqrt{x} - x^{3} 2 \frac{1}{2 \sqrt{x}}}{\left(2 \sqrt{x}\right)^{2}} - \left(\frac{3 x^{2}}{2 \sqrt{x}} + \sqrt{x} 6 x\right)\right) \left(x^{3}\right)^{2} - \left(\frac{x^{3}}{2 \sqrt{x}} - \sqrt{x} 3 x^{2}\right) 2 x^{3} 3 x^{2}}{\left(\left(x^{3}\right)^{2}\right)^{2}} \]

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

xf(x)f'(x)
0.55.6568542-28.284271
11-2.5
20.1767767-0.22097087
30.06415003-0.053458358

Questions

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

The derivative of xx3\frac{\sqrt{x}}{x^{3}} is x32x−x3x2(x3)2\frac{\frac{x^{3}}{2 \sqrt{x}} - \sqrt{x} 3 x^{2}}{\left(x^{3}\right)^{2}}.

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

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

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

The quotient rule.

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

-2.5.

Related derivatives

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