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Derivative of tan⁡(x)arctan⁡(x)\tan\left(x\right) \arctan\left(x\right)

The derivative of tan⁡(x)arctan⁡(x)\tan\left(x\right) \arctan\left(x\right) is arctan⁡(x)cos⁡(x)2+tan⁡(x)11+x2\frac{\arctan\left(x\right)}{\cos\left(x\right)^{2}} + \tan\left(x\right) \frac{1}{1 + x^{2}}.

f(x) =
ddx[tan⁡(x)arctan⁡(x)]=arctan⁡(x)cos⁡(x)2+tan⁡(x)11+x2\frac{d}{dx}\left[\tan\left(x\right) \arctan\left(x\right)\right] = \frac{\arctan\left(x\right)}{\cos\left(x\right)^{2}} + \tan\left(x\right) \frac{1}{1 + x^{2}}

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Graph of f(x) = tan(x) * atan(x) and its derivative-3-2-1123-20-101020slope 3.469f(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) = 3.4691.

How to work it out

  1. 1

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

    f(x)=tan⁡(x)arctan⁡(x)f(x) = \tan\left(x\right) \arctan\left(x\right)
  2. 2

    Product rule: (uv)′=u′v+uv′(uv)' = u'v + uv'. 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)f'(x) = \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)
  5. 5

    Simplify:

    f′(x)=arctan⁡(x)cos⁡(x)2+tan⁡(x)11+x2f'(x) = \frac{\arctan\left(x\right)}{\cos\left(x\right)^{2}} + \tan\left(x\right) \frac{1}{1 + x^{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)2f''(x) = \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}} + \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}}
LaTeX source of the formulas on this page
\[ \frac{d}{dx}\left[\tan\left(x\right) \arctan\left(x\right)\right] = \frac{\arctan\left(x\right)}{\cos\left(x\right)^{2}} + \tan\left(x\right) \frac{1}{1 + x^{2}} \]
\[ f(x) = \tan\left(x\right) \arctan\left(x\right) \]
\[ u = \tan\left(x\right),\quad v = \arctan\left(x\right) \]
\[ u' = \frac{1}{\cos\left(x\right)^{2}},\quad v' = \frac{1}{1 + x^{2}} \]
\[ f'(x) = \left(\frac{1}{\cos\left(x\right)^{2}}\right)\left(\arctan\left(x\right)\right) + \left(\tan\left(x\right)\right)\left(\frac{1}{1 + x^{2}}\right) \]
\[ f'(x) = \frac{\arctan\left(x\right)}{\cos\left(x\right)^{2}} + \tan\left(x\right) \frac{1}{1 + x^{2}} \]
\[ f''(x) = \frac{\frac{\cos\left(x\right)^{2}}{1 + x^{2}} - \arctan\left(x\right) 2 \cos\left(x\right) -\sin\left(x\right)}{\left(\cos\left(x\right)^{2}\right)^{2}} + \frac{\frac{1}{1 + x^{2}}}{\cos\left(x\right)^{2}} + \tan\left(x\right) \frac{-\left(2 x\right)}{\left(1 + x^{2}\right)^{2}} \]

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

xf(x)f'(x)
-2-2.4191641-5.9561107
-11.2231852-3.4691001
0.50.253291841.0390636
11.22318523.4691001
2-2.41916415.9561107
3-0.178047161.2601711

Questions

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

The derivative of tan⁡(x)arctan⁡(x)\tan\left(x\right) \arctan\left(x\right) is arctan⁡(x)cos⁡(x)2+tan⁡(x)11+x2\frac{\arctan\left(x\right)}{\cos\left(x\right)^{2}} + \tan\left(x\right) \frac{1}{1 + x^{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.

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

The product rule.

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

3.4691001.

Related derivatives

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