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

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

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

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Graph of f(x) = atan(x) / sin(x) and its derivative-3-2-1123-2246slope -0.005108f(x)f'(x)tangent at x = 1
Graph of f(x) = atan(x) / sin(x) and its derivative. The dashed line is the tangent at x = 1; its slope is f'(1) = -0.0051082.

How to work it out

  1. 1

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

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

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

    Differentiate each piece:

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

    Put them in the rule:

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

    Simplify:

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

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

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

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

xf(x)f'(x)
-21.217587-0.77718792
-10.933363330.0051081515
0.50.96708993-0.10158252
10.93336333-0.0051081515
21.2175870.77718792
38.850947462.800248

Questions

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

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

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

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

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

The quotient rule.

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

-0.0051081515.

Related derivatives

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