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Derivative of 1xsin⁡(x)\frac{\frac{1}{x}}{\sin\left(x\right)}

The derivative of 1xsin⁡(x)\frac{\frac{1}{x}}{\sin\left(x\right)} is −1x2sin⁡(x)−cos⁡(x)xsin⁡(x)2\frac{\frac{-1}{x^{2}} \sin\left(x\right) - \frac{\cos\left(x\right)}{x}}{\sin\left(x\right)^{2}}.

f(x) =
ddx[1xsin⁡(x)]=−1x2sin⁡(x)−cos⁡(x)xsin⁡(x)2\frac{d}{dx}\left[\frac{\frac{1}{x}}{\sin\left(x\right)}\right] = \frac{\frac{-1}{x^{2}} \sin\left(x\right) - \frac{\cos\left(x\right)}{x}}{\sin\left(x\right)^{2}}

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Graph of f(x) = 1 / x / sin(x) and its derivative-3-2-1123-2020406080slope -1.951f(x)f'(x)tangent at x = 1
Graph of f(x) = 1 / x / sin(x) and its derivative. The dashed line is the tangent at x = 1; its slope is f'(1) = -1.9515.

How to work it out

  1. 1

    We differentiate f(x)=1xsin⁡(x)f(x) = \frac{\frac{1}{x}}{\sin\left(x\right)} with respect to x. The rule that applies is the quotient rule.

    f(x)=1xsin⁡(x)f(x) = \frac{\frac{1}{x}}{\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=1xu = \frac{1}{x} and v=sin⁡(x)v = \sin\left(x\right).

    u=1x,v=sin⁡(x)u = \frac{1}{x},\quad v = \sin\left(x\right)
  3. 3

    Differentiate each piece:

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

    Put them in the rule:

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

    Simplify:

    f′(x)=−1x2sin⁡(x)−cos⁡(x)xsin⁡(x)2f'(x) = \frac{\frac{-1}{x^{2}} \sin\left(x\right) - \frac{\cos\left(x\right)}{x}}{\sin\left(x\right)^{2}}
  6. 6

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

    f′′(x)=(−((−2)x)(x2)2sin⁡(x)+−1x2cos⁡(x)−−(sin⁡(x)x)−cos⁡(x)x2)sin⁡(x)2−(−1x2sin⁡(x)−cos⁡(x)x)2sin⁡(x)cos⁡(x)(sin⁡(x)2)2f''(x) = \frac{\left(\frac{-\left(\left(-2\right) x\right)}{\left(x^{2}\right)^{2}} \sin\left(x\right) + \frac{-1}{x^{2}} \cos\left(x\right) - \frac{-\left(\sin\left(x\right) x\right) - \cos\left(x\right)}{x^{2}}\right) \sin\left(x\right)^{2} - \left(\frac{-1}{x^{2}} \sin\left(x\right) - \frac{\cos\left(x\right)}{x}\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
\[ \frac{d}{dx}\left[\frac{\frac{1}{x}}{\sin\left(x\right)}\right] = \frac{\frac{-1}{x^{2}} \sin\left(x\right) - \frac{\cos\left(x\right)}{x}}{\sin\left(x\right)^{2}} \]
\[ f(x) = \frac{\frac{1}{x}}{\sin\left(x\right)} \]
\[ u = \frac{1}{x},\quad v = \sin\left(x\right) \]
\[ u' = \frac{-1}{x^{2}},\quad v' = \cos\left(x\right) \]
\[ f'(x) = \frac{\left(\frac{-1}{x^{2}}\right)\left(\sin\left(x\right)\right) - \left(\frac{1}{x}\right)\left(\cos\left(x\right)\right)}{\left(\sin\left(x\right)\right)^{2}} \]
\[ f'(x) = \frac{\frac{-1}{x^{2}} \sin\left(x\right) - \frac{\cos\left(x\right)}{x}}{\sin\left(x\right)^{2}} \]
\[ f''(x) = \frac{\left(\frac{-\left(\left(-2\right) x\right)}{\left(x^{2}\right)^{2}} \sin\left(x\right) + \frac{-1}{x^{2}} \cos\left(x\right) - \frac{-\left(\sin\left(x\right) x\right) - \cos\left(x\right)}{x^{2}}\right) \sin\left(x\right)^{2} - \left(\frac{-1}{x^{2}} \sin\left(x\right) - \frac{\cos\left(x\right)}{x}\right) 2 \sin\left(x\right) \cos\left(x\right)}{\left(\sin\left(x\right)^{2}\right)^{2}} \]

Values of f and f' for 1 / x / sin(x)

xf(x)f'(x)
-20.549875090.023283056
-11.18839511.9514548
0.54.1716593-15.97949
11.1883951-1.9514548
20.54987509-0.023283056
32.362055815.783066

Questions

What is the derivative of 1 / x / sin(x)?

The derivative of 1xsin⁡(x)\frac{\frac{1}{x}}{\sin\left(x\right)} is −1x2sin⁡(x)−cos⁡(x)xsin⁡(x)2\frac{\frac{-1}{x^{2}} \sin\left(x\right) - \frac{\cos\left(x\right)}{x}}{\sin\left(x\right)^{2}}.

What is the second derivative of 1 / x / sin(x)?

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

Which rule do you use for 1 / x / sin(x)?

The quotient rule.

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

-1.9514548.

Related derivatives

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