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Derivative of arctan⁡(x)tan⁡(x)\frac{\arctan\left(x\right)}{\tan\left(x\right)}

The derivative of arctan⁡(x)tan⁡(x)\frac{\arctan\left(x\right)}{\tan\left(x\right)} is tan⁡(x)1+x2−arctan⁡(x)1cos⁡(x)2tan⁡(x)2\frac{\frac{\tan\left(x\right)}{1 + x^{2}} - \arctan\left(x\right) \frac{1}{\cos\left(x\right)^{2}}}{\tan\left(x\right)^{2}}.

f(x) =
ddx[arctan⁡(x)tan⁡(x)]=tan⁡(x)1+x2−arctan⁡(x)1cos⁡(x)2tan⁡(x)2\frac{d}{dx}\left[\frac{\arctan\left(x\right)}{\tan\left(x\right)}\right] = \frac{\frac{\tan\left(x\right)}{1 + x^{2}} - \arctan\left(x\right) \frac{1}{\cos\left(x\right)^{2}}}{\tan\left(x\right)^{2}}

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Graph of f(x) = atan(x) / tan(x) and its derivative-3-2-1123-6-4-224slope -0.7882f(x)f'(x)tangent at x = 1
Graph of f(x) = atan(x) / tan(x) and its derivative. The dashed line is the tangent at x = 1; its slope is f'(1) = -0.78816.

How to work it out

  1. 1

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

    f(x)=arctan⁡(x)tan⁡(x)f(x) = \frac{\arctan\left(x\right)}{\tan\left(x\right)}
  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=tan⁡(x)v = \tan\left(x\right).

    u=arctan⁡(x),v=tan⁡(x)u = \arctan\left(x\right),\quad v = \tan\left(x\right)
  3. 3

    Differentiate each piece:

    u′=11+x2,v′=1cos⁡(x)2u' = \frac{1}{1 + x^{2}},\quad v' = \frac{1}{\cos\left(x\right)^{2}}
  4. 4

    Put them in the rule:

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

    Simplify:

    f′(x)=tan⁡(x)1+x2−arctan⁡(x)1cos⁡(x)2tan⁡(x)2f'(x) = \frac{\frac{\tan\left(x\right)}{1 + x^{2}} - \arctan\left(x\right) \frac{1}{\cos\left(x\right)^{2}}}{\tan\left(x\right)^{2}}
  6. 6

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

    f′′(x)=(1+x2cos⁡(x)2−tan⁡(x)2x(1+x2)2−(1cos⁡(x)21+x2+arctan⁡(x)−(2cos⁡(x)−sin⁡(x))(cos⁡(x)2)2))tan⁡(x)2−(tan⁡(x)1+x2−arctan⁡(x)1cos⁡(x)2)2tan⁡(x)1cos⁡(x)2(tan⁡(x)2)2f''(x) = \frac{\left(\frac{\frac{1 + x^{2}}{\cos\left(x\right)^{2}} - \tan\left(x\right) 2 x}{\left(1 + x^{2}\right)^{2}} - \left(\frac{\frac{1}{\cos\left(x\right)^{2}}}{1 + x^{2}} + \arctan\left(x\right) \frac{-\left(2 \cos\left(x\right) -\sin\left(x\right)\right)}{\left(\cos\left(x\right)^{2}\right)^{2}}\right)\right) \tan\left(x\right)^{2} - \left(\frac{\tan\left(x\right)}{1 + x^{2}} - \arctan\left(x\right) \frac{1}{\cos\left(x\right)^{2}}\right) 2 \tan\left(x\right) \frac{1}{\cos\left(x\right)^{2}}}{\left(\tan\left(x\right)^{2}\right)^{2}}
LaTeX source of the formulas on this page
\[ \frac{d}{dx}\left[\frac{\arctan\left(x\right)}{\tan\left(x\right)}\right] = \frac{\frac{\tan\left(x\right)}{1 + x^{2}} - \arctan\left(x\right) \frac{1}{\cos\left(x\right)^{2}}}{\tan\left(x\right)^{2}} \]
\[ f(x) = \frac{\arctan\left(x\right)}{\tan\left(x\right)} \]
\[ u = \arctan\left(x\right),\quad v = \tan\left(x\right) \]
\[ u' = \frac{1}{1 + x^{2}},\quad v' = \frac{1}{\cos\left(x\right)^{2}} \]
\[ f'(x) = \frac{\left(\frac{1}{1 + x^{2}}\right)\left(\tan\left(x\right)\right) - \left(\arctan\left(x\right)\right)\left(\frac{1}{\cos\left(x\right)^{2}}\right)}{\left(\tan\left(x\right)\right)^{2}} \]
\[ f'(x) = \frac{\frac{\tan\left(x\right)}{1 + x^{2}} - \arctan\left(x\right) \frac{1}{\cos\left(x\right)^{2}}}{\tan\left(x\right)^{2}} \]
\[ f''(x) = \frac{\left(\frac{\frac{1 + x^{2}}{\cos\left(x\right)^{2}} - \tan\left(x\right) 2 x}{\left(1 + x^{2}\right)^{2}} - \left(\frac{\frac{1}{\cos\left(x\right)^{2}}}{1 + x^{2}} + \arctan\left(x\right) \frac{-\left(2 \cos\left(x\right) -\sin\left(x\right)\right)}{\left(\cos\left(x\right)^{2}\right)^{2}}\right)\right) \tan\left(x\right)^{2} - \left(\frac{\tan\left(x\right)}{1 + x^{2}} - \arctan\left(x\right) \frac{1}{\cos\left(x\right)^{2}}\right) 2 \tan\left(x\right) \frac{1}{\cos\left(x\right)^{2}}}{\left(\tan\left(x\right)^{2}\right)^{2}} \]

Values of f and f' for atan(x) / tan(x)

xf(x)f'(x)
-2-0.506694971.430573
-10.504298360.78815811
0.50.84870126-0.55279466
10.50429836-0.78815811
2-0.50669497-1.430573
3-8.7623715-63.42082

Questions

What is the derivative of atan(x) / tan(x)?

The derivative of arctan⁡(x)tan⁡(x)\frac{\arctan\left(x\right)}{\tan\left(x\right)} is tan⁡(x)1+x2−arctan⁡(x)1cos⁡(x)2tan⁡(x)2\frac{\frac{\tan\left(x\right)}{1 + x^{2}} - \arctan\left(x\right) \frac{1}{\cos\left(x\right)^{2}}}{\tan\left(x\right)^{2}}.

What is the second derivative of atan(x) / tan(x)?

((((1 + x2x^{2}) / cos(x)^2 - tan(x) * 2x) / (1 + x2x^{2})^2 - (1 / cos(x)^2 / (1 + x2x^{2}) + atan(x) * -(2cos(x) * -sin(x)) / (cos(x)^2)^2)) * tan(x)^2 - (tan(x) / (1 + x2x^{2}) - atan(x) * 1 / cos(x)^2) * 2tan(x) * 1 / cos(x)^2) / (tan(x)^2)^2.

Which rule do you use for atan(x) / tan(x)?

The quotient rule.

What is the slope of atan(x) / tan(x) at x = 1?

-0.78815811.

Related derivatives

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