WordGPT

Derivative of xx2\frac{\sqrt{x}}{x^{2}}

The derivative of xx2\frac{\sqrt{x}}{x^{2}} is x22x−x2x(x2)2\frac{\frac{x^{2}}{2 \sqrt{x}} - \sqrt{x} 2 x}{\left(x^{2}\right)^{2}}.

f(x) =
ddx[xx2]=x22x−x2x(x2)2\frac{d}{dx}\left[\frac{\sqrt{x}}{x^{2}}\right] = \frac{\frac{x^{2}}{2 \sqrt{x}} - \sqrt{x} 2 x}{\left(x^{2}\right)^{2}}

Type any function of x; the derivative is computed symbolically in your browser.

Explain it step by step in chatMake a short video of it
Graph of f(x) = sqrt(x) / x^2 and its derivative-3-2-1123-20-15-10-5510slope -1.5f(x)f'(x)tangent at x = 1
Graph of f(x) = sqrt(x) / x^2 and its derivative. The dashed line is the tangent at x = 1; its slope is f'(1) = -1.5.

How to work it out

  1. 1

    We differentiate f(x)=xx2f(x) = \frac{\sqrt{x}}{x^{2}} with respect to x. The rule that applies is the quotient rule.

    f(x)=xx2f(x) = \frac{\sqrt{x}}{x^{2}}
  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=x2v = x^{2}.

    u=x,v=x2u = \sqrt{x},\quad v = x^{2}
  3. 3

    Differentiate each piece:

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

    Put them in the rule:

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

    Simplify:

    f′(x)=x22x−x2x(x2)2f'(x) = \frac{\frac{x^{2}}{2 \sqrt{x}} - \sqrt{x} 2 x}{\left(x^{2}\right)^{2}}
  6. 6

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

    f′′(x)=(2x2x−x2212x(2x)2−(2x2x+2x))(x2)2−(x22x−x2x)2x22x((x2)2)2f''(x) = \frac{\left(\frac{2 x 2 \sqrt{x} - x^{2} 2 \frac{1}{2 \sqrt{x}}}{\left(2 \sqrt{x}\right)^{2}} - \left(\frac{2 x}{2 \sqrt{x}} + 2 \sqrt{x}\right)\right) \left(x^{2}\right)^{2} - \left(\frac{x^{2}}{2 \sqrt{x}} - \sqrt{x} 2 x\right) 2 x^{2} 2 x}{\left(\left(x^{2}\right)^{2}\right)^{2}}
LaTeX source of the formulas on this page
\[ \frac{d}{dx}\left[\frac{\sqrt{x}}{x^{2}}\right] = \frac{\frac{x^{2}}{2 \sqrt{x}} - \sqrt{x} 2 x}{\left(x^{2}\right)^{2}} \]
\[ f(x) = \frac{\sqrt{x}}{x^{2}} \]
\[ u = \sqrt{x},\quad v = x^{2} \]
\[ u' = \frac{1}{2 \sqrt{x}},\quad v' = 2 x \]
\[ f'(x) = \frac{\left(\frac{1}{2 \sqrt{x}}\right)\left(x^{2}\right) - \left(\sqrt{x}\right)\left(2 x\right)}{\left(x^{2}\right)^{2}} \]
\[ f'(x) = \frac{\frac{x^{2}}{2 \sqrt{x}} - \sqrt{x} 2 x}{\left(x^{2}\right)^{2}} \]
\[ f''(x) = \frac{\left(\frac{2 x 2 \sqrt{x} - x^{2} 2 \frac{1}{2 \sqrt{x}}}{\left(2 \sqrt{x}\right)^{2}} - \left(\frac{2 x}{2 \sqrt{x}} + 2 \sqrt{x}\right)\right) \left(x^{2}\right)^{2} - \left(\frac{x^{2}}{2 \sqrt{x}} - \sqrt{x} 2 x\right) 2 x^{2} 2 x}{\left(\left(x^{2}\right)^{2}\right)^{2}} \]

Values of f and f' for sqrt(x) / x2x^{2}

xf(x)f'(x)
0.52.8284271-8.4852814
11-1.5
20.35355339-0.26516504
30.19245009-0.096225045

Questions

What is the derivative of sqrt(x) / x2x^{2}?

The derivative of xx2\frac{\sqrt{x}}{x^{2}} is x22x−x2x(x2)2\frac{\frac{x^{2}}{2 \sqrt{x}} - \sqrt{x} 2 x}{\left(x^{2}\right)^{2}}.

What is the second derivative of sqrt(x) / x2x^{2}?

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

Which rule do you use for sqrt(x) / x2x^{2}?

The quotient rule.

What is the slope of sqrt(x) / x2x^{2} at x = 1?

-1.5.

Related derivatives

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