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

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

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

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Graph of f(x) = cos(x) * atan(x) and its derivative-3-2-1123-1-0.50.511.52slope -0.3907f(x)f'(x)tangent at x = 1
Graph of f(x) = cos(x) * atan(x) and its derivative. The dashed line is the tangent at x = 1; its slope is f'(1) = -0.39074.

How to work it out

  1. 1

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

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

    Product rule: (uv)′=u′v+uv′(uv)' = u'v + uv'. Let u=cos⁡(x)u = \cos\left(x\right) and v=arctan⁡(x)v = \arctan\left(x\right).

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

    Differentiate each piece:

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

    Put them in the rule:

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

    Simplify:

    f′(x)=−(sin⁡(x)arctan⁡(x))+cos⁡(x)11+x2f'(x) = -\left(\sin\left(x\right) \arctan\left(x\right)\right) + \cos\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)arctan⁡(x)+sin⁡(x)11+x2)+−(sin⁡(x)11+x2)+cos⁡(x)−(2x)(1+x2)2f''(x) = -\left(\cos\left(x\right) \arctan\left(x\right) + \sin\left(x\right) \frac{1}{1 + x^{2}}\right) + -\left(\sin\left(x\right) \frac{1}{1 + x^{2}}\right) + \cos\left(x\right) \frac{-\left(2 x\right)}{\left(1 + x^{2}\right)^{2}}
LaTeX source of the formulas on this page
\[ \frac{d}{dx}\left[\cos\left(x\right) \arctan\left(x\right)\right] = -\left(\sin\left(x\right) \arctan\left(x\right)\right) + \cos\left(x\right) \frac{1}{1 + x^{2}} \]
\[ f(x) = \cos\left(x\right) \arctan\left(x\right) \]
\[ u = \cos\left(x\right),\quad v = \arctan\left(x\right) \]
\[ u' = -\sin\left(x\right),\quad v' = \frac{1}{1 + x^{2}} \]
\[ f'(x) = \left(-\sin\left(x\right)\right)\left(\arctan\left(x\right)\right) + \left(\cos\left(x\right)\right)\left(\frac{1}{1 + x^{2}}\right) \]
\[ f'(x) = -\left(\sin\left(x\right) \arctan\left(x\right)\right) + \cos\left(x\right) \frac{1}{1 + x^{2}} \]
\[ f''(x) = -\left(\cos\left(x\right) \arctan\left(x\right) + \sin\left(x\right) \frac{1}{1 + x^{2}}\right) + -\left(\sin\left(x\right) \frac{1}{1 + x^{2}}\right) + \cos\left(x\right) \frac{-\left(2 x\right)}{\left(1 + x^{2}\right)^{2}} \]

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

xf(x)f'(x)
-20.46073644-1.0899568
-1-0.42435244-0.39073861
0.50.406889060.47978154
10.42435244-0.39073861
2-0.46073644-1.0899568
3-1.2365459-0.2752646

Questions

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

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

What is the second derivative of cos(x) * atan(x)?

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

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

The product rule.

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

-0.39073861.

Related derivatives

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