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

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

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

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Graph of f(x) = sin(x) / atan(x) and its derivative-3-2-1123-0.50.51slope 0.005864f(x)f'(x)tangent at x = 1
Graph of f(x) = sin(x) / atan(x) and its derivative. The dashed line is the tangent at x = 1; its slope is f'(1) = 0.0058636.

How to work it out

  1. 1

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

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

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

    Differentiate each piece:

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

    Put them in the rule:

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

    Simplify:

    f′(x)=cos⁡(x)arctan⁡(x)−sin⁡(x)11+x2arctan⁡(x)2f'(x) = \frac{\cos\left(x\right) \arctan\left(x\right) - \sin\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)=(−(sin⁡(x)arctan⁡(x))+cos⁡(x)11+x2−(cos⁡(x)11+x2+sin⁡(x)−(2x)(1+x2)2))arctan⁡(x)2−(cos⁡(x)arctan⁡(x)−sin⁡(x)11+x2)2arctan⁡(x)11+x2(arctan⁡(x)2)2f''(x) = \frac{\left(-\left(\sin\left(x\right) \arctan\left(x\right)\right) + \cos\left(x\right) \frac{1}{1 + x^{2}} - \left(\cos\left(x\right) \frac{1}{1 + x^{2}} + \sin\left(x\right) \frac{-\left(2 x\right)}{\left(1 + x^{2}\right)^{2}}\right)\right) \arctan\left(x\right)^{2} - \left(\cos\left(x\right) \arctan\left(x\right) - \sin\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
\[ \frac{d}{dx}\left[\frac{\sin\left(x\right)}{\arctan\left(x\right)}\right] = \frac{\cos\left(x\right) \arctan\left(x\right) - \sin\left(x\right) \frac{1}{1 + x^{2}}}{\arctan\left(x\right)^{2}} \]
\[ f(x) = \frac{\sin\left(x\right)}{\arctan\left(x\right)} \]
\[ u = \sin\left(x\right),\quad v = \arctan\left(x\right) \]
\[ u' = \cos\left(x\right),\quad v' = \frac{1}{1 + x^{2}} \]
\[ f'(x) = \frac{\left(\cos\left(x\right)\right)\left(\arctan\left(x\right)\right) - \left(\sin\left(x\right)\right)\left(\frac{1}{1 + x^{2}}\right)}{\left(\arctan\left(x\right)\right)^{2}} \]
\[ f'(x) = \frac{\cos\left(x\right) \arctan\left(x\right) - \sin\left(x\right) \frac{1}{1 + x^{2}}}{\arctan\left(x\right)^{2}} \]
\[ f''(x) = \frac{\left(-\left(\sin\left(x\right) \arctan\left(x\right)\right) + \cos\left(x\right) \frac{1}{1 + x^{2}} - \left(\cos\left(x\right) \frac{1}{1 + x^{2}} + \sin\left(x\right) \frac{-\left(2 x\right)}{\left(1 + x^{2}\right)^{2}}\right)\right) \arctan\left(x\right)^{2} - \left(\cos\left(x\right) \arctan\left(x\right) - \sin\left(x\right) \frac{1}{1 + x^{2}}\right) 2 \arctan\left(x\right) \frac{1}{1 + x^{2}}}{\left(\arctan\left(x\right)^{2}\right)^{2}} \]

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

xf(x)f'(x)
-20.821296550.52423504
-11.0713941-0.0058635725
0.51.034030.10861387
11.07139410.0058635725
20.82129655-0.52423504
30.11298226-0.80164454

Questions

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

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

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

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

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

The quotient rule.

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

0.0058635725.

Related derivatives

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