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Derivative of arctan⁡(x)x4\frac{\arctan\left(x\right)}{x^{4}}

The derivative of arctan⁡(x)x4\frac{\arctan\left(x\right)}{x^{4}} is x41+x2−arctan⁡(x)4x3(x4)2\frac{\frac{x^{4}}{1 + x^{2}} - \arctan\left(x\right) 4 x^{3}}{\left(x^{4}\right)^{2}}.

f(x) =
ddx[arctan⁡(x)x4]=x41+x2−arctan⁡(x)4x3(x4)2\frac{d}{dx}\left[\frac{\arctan\left(x\right)}{x^{4}}\right] = \frac{\frac{x^{4}}{1 + x^{2}} - \arctan\left(x\right) 4 x^{3}}{\left(x^{4}\right)^{2}}

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Graph of f(x) = atan(x) / x^4 and its derivative-3-2-1123-1000-800-600-400-200slope -2.642f(x)f'(x)tangent at x = 1
Graph of f(x) = atan(x) / x^4 and its derivative. The dashed line is the tangent at x = 1; its slope is f'(1) = -2.6416.

How to work it out

  1. 1

    We differentiate f(x)=arctan⁡(x)x4f(x) = \frac{\arctan\left(x\right)}{x^{4}} with respect to x. The rule that applies is the quotient rule.

    f(x)=arctan⁡(x)x4f(x) = \frac{\arctan\left(x\right)}{x^{4}}
  2. 2

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

    u=arctan⁡(x),v=x4u = \arctan\left(x\right),\quad v = x^{4}
  3. 3

    Differentiate each piece:

    u′=11+x2,v′=4x3u' = \frac{1}{1 + x^{2}},\quad v' = 4 x^{3}
  4. 4

    Put them in the rule:

    f′(x)=(11+x2)(x4)−(arctan⁡(x))(4x3)(x4)2f'(x) = \frac{\left(\frac{1}{1 + x^{2}}\right)\left(x^{4}\right) - \left(\arctan\left(x\right)\right)\left(4 x^{3}\right)}{\left(x^{4}\right)^{2}}
  5. 5

    Simplify:

    f′(x)=x41+x2−arctan⁡(x)4x3(x4)2f'(x) = \frac{\frac{x^{4}}{1 + x^{2}} - \arctan\left(x\right) 4 x^{3}}{\left(x^{4}\right)^{2}}
  6. 6

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

    f′′(x)=(4x3(1+x2)−x42x(1+x2)2−(4x31+x2+arctan⁡(x)12x2))(x4)2−(x41+x2−arctan⁡(x)4x3)2x44x3((x4)2)2f''(x) = \frac{\left(\frac{4 x^{3} \left(1 + x^{2}\right) - x^{4} 2 x}{\left(1 + x^{2}\right)^{2}} - \left(\frac{4 x^{3}}{1 + x^{2}} + \arctan\left(x\right) 12 x^{2}\right)\right) \left(x^{4}\right)^{2} - \left(\frac{x^{4}}{1 + x^{2}} - \arctan\left(x\right) 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
\[ \frac{d}{dx}\left[\frac{\arctan\left(x\right)}{x^{4}}\right] = \frac{\frac{x^{4}}{1 + x^{2}} - \arctan\left(x\right) 4 x^{3}}{\left(x^{4}\right)^{2}} \]
\[ f(x) = \frac{\arctan\left(x\right)}{x^{4}} \]
\[ u = \arctan\left(x\right),\quad v = x^{4} \]
\[ u' = \frac{1}{1 + x^{2}},\quad v' = 4 x^{3} \]
\[ f'(x) = \frac{\left(\frac{1}{1 + x^{2}}\right)\left(x^{4}\right) - \left(\arctan\left(x\right)\right)\left(4 x^{3}\right)}{\left(x^{4}\right)^{2}} \]
\[ f'(x) = \frac{\frac{x^{4}}{1 + x^{2}} - \arctan\left(x\right) 4 x^{3}}{\left(x^{4}\right)^{2}} \]
\[ f''(x) = \frac{\left(\frac{4 x^{3} \left(1 + x^{2}\right) - x^{4} 2 x}{\left(1 + x^{2}\right)^{2}} - \left(\frac{4 x^{3}}{1 + x^{2}} + \arctan\left(x\right) 12 x^{2}\right)\right) \left(x^{4}\right)^{2} - \left(\frac{x^{4}}{1 + x^{2}} - \arctan\left(x\right) 4 x^{3}\right) 2 x^{4} 4 x^{3}}{\left(\left(x^{4}\right)^{2}\right)^{2}} \]

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

xf(x)f'(x)
-2-0.069196795-0.12589359
-1-0.78539816-2.6415927
0.57.4183617-46.546894
10.78539816-2.6415927
20.069196795-0.12589359
30.015420318-0.019325856

Questions

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

The derivative of arctan⁡(x)x4\frac{\arctan\left(x\right)}{x^{4}} is x41+x2−arctan⁡(x)4x3(x4)2\frac{\frac{x^{4}}{1 + x^{2}} - \arctan\left(x\right) 4 x^{3}}{\left(x^{4}\right)^{2}}.

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

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

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

The quotient rule.

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

-2.6415927.

Related derivatives

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