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

The derivative of tan⁡(x)arctan⁡(x)\frac{\tan\left(x\right)}{\arctan\left(x\right)} is arctan⁡(x)cos⁡(x)2−tan⁡(x)11+x2arctan⁡(x)2\frac{\frac{\arctan\left(x\right)}{\cos\left(x\right)^{2}} - \tan\left(x\right) \frac{1}{1 + x^{2}}}{\arctan\left(x\right)^{2}}.

f(x) =
ddx[tan⁡(x)arctan⁡(x)]=arctan⁡(x)cos⁡(x)2−tan⁡(x)11+x2arctan⁡(x)2\frac{d}{dx}\left[\frac{\tan\left(x\right)}{\arctan\left(x\right)}\right] = \frac{\frac{\arctan\left(x\right)}{\cos\left(x\right)^{2}} - \tan\left(x\right) \frac{1}{1 + x^{2}}}{\arctan\left(x\right)^{2}}

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Graph of f(x) = tan(x) / atan(x) and its derivative-3-2-1123-20-101020slope 3.099f(x)f'(x)tangent at x = 1
Graph of f(x) = tan(x) / atan(x) and its derivative. The dashed line is the tangent at x = 1; its slope is f'(1) = 3.0991.

How to work it out

  1. 1

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

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

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

    Differentiate each piece:

    u′=1cos⁡(x)2,v′=11+x2u' = \frac{1}{\cos\left(x\right)^{2}},\quad v' = \frac{1}{1 + x^{2}}
  4. 4

    Put them in the rule:

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

    Simplify:

    f′(x)=arctan⁡(x)cos⁡(x)2−tan⁡(x)11+x2arctan⁡(x)2f'(x) = \frac{\frac{\arctan\left(x\right)}{\cos\left(x\right)^{2}} - \tan\left(x\right) \frac{1}{1 + x^{2}}}{\arctan\left(x\right)^{2}}
  6. 6

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

    f′′(x)=(cos⁡(x)21+x2−arctan⁡(x)2cos⁡(x)−sin⁡(x)(cos⁡(x)2)2−(11+x2cos⁡(x)2+tan⁡(x)−(2x)(1+x2)2))arctan⁡(x)2−(arctan⁡(x)cos⁡(x)2−tan⁡(x)11+x2)2arctan⁡(x)11+x2(arctan⁡(x)2)2f''(x) = \frac{\left(\frac{\frac{\cos\left(x\right)^{2}}{1 + x^{2}} - \arctan\left(x\right) 2 \cos\left(x\right) -\sin\left(x\right)}{\left(\cos\left(x\right)^{2}\right)^{2}} - \left(\frac{\frac{1}{1 + x^{2}}}{\cos\left(x\right)^{2}} + \tan\left(x\right) \frac{-\left(2 x\right)}{\left(1 + x^{2}\right)^{2}}\right)\right) \arctan\left(x\right)^{2} - \left(\frac{\arctan\left(x\right)}{\cos\left(x\right)^{2}} - \tan\left(x\right) \frac{1}{1 + x^{2}}\right) 2 \arctan\left(x\right) \frac{1}{1 + x^{2}}}{\left(\arctan\left(x\right)^{2}\right)^{2}}
LaTeX source of the formulas on this page
\[ \frac{d}{dx}\left[\frac{\tan\left(x\right)}{\arctan\left(x\right)}\right] = \frac{\frac{\arctan\left(x\right)}{\cos\left(x\right)^{2}} - \tan\left(x\right) \frac{1}{1 + x^{2}}}{\arctan\left(x\right)^{2}} \]
\[ f(x) = \frac{\tan\left(x\right)}{\arctan\left(x\right)} \]
\[ u = \tan\left(x\right),\quad v = \arctan\left(x\right) \]
\[ u' = \frac{1}{\cos\left(x\right)^{2}},\quad v' = \frac{1}{1 + x^{2}} \]
\[ f'(x) = \frac{\left(\frac{1}{\cos\left(x\right)^{2}}\right)\left(\arctan\left(x\right)\right) - \left(\tan\left(x\right)\right)\left(\frac{1}{1 + x^{2}}\right)}{\left(\arctan\left(x\right)\right)^{2}} \]
\[ f'(x) = \frac{\frac{\arctan\left(x\right)}{\cos\left(x\right)^{2}} - \tan\left(x\right) \frac{1}{1 + x^{2}}}{\arctan\left(x\right)^{2}} \]
\[ f''(x) = \frac{\left(\frac{\frac{\cos\left(x\right)^{2}}{1 + x^{2}} - \arctan\left(x\right) 2 \cos\left(x\right) -\sin\left(x\right)}{\left(\cos\left(x\right)^{2}\right)^{2}} - \left(\frac{\frac{1}{1 + x^{2}}}{\cos\left(x\right)^{2}} + \tan\left(x\right) \frac{-\left(2 x\right)}{\left(1 + x^{2}\right)^{2}}\right)\right) \arctan\left(x\right)^{2} - \left(\frac{\arctan\left(x\right)}{\cos\left(x\right)^{2}} - \tan\left(x\right) \frac{1}{1 + x^{2}}\right) 2 \arctan\left(x\right) \frac{1}{1 + x^{2}}}{\left(\arctan\left(x\right)^{2}\right)^{2}} \]

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

xf(x)f'(x)
-2-1.9735739-5.5720735
-11.9829531-3.0991189
0.51.17827090.76745717
11.98295313.0991189
2-1.97357395.5720735
3-0.114124350.82601613

Questions

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

The derivative of tan⁡(x)arctan⁡(x)\frac{\tan\left(x\right)}{\arctan\left(x\right)} is arctan⁡(x)cos⁡(x)2−tan⁡(x)11+x2arctan⁡(x)2\frac{\frac{\arctan\left(x\right)}{\cos\left(x\right)^{2}} - \tan\left(x\right) \frac{1}{1 + x^{2}}}{\arctan\left(x\right)^{2}}.

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

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

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

The quotient rule.

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

3.0991189.

Related derivatives

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