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Derivative of 1xtan⁡(x)\frac{\frac{1}{x}}{\tan\left(x\right)}

The derivative of 1xtan⁡(x)\frac{\frac{1}{x}}{\tan\left(x\right)} is −1x2tan⁡(x)−1cos⁡(x)2xtan⁡(x)2\frac{\frac{-1}{x^{2}} \tan\left(x\right) - \frac{\frac{1}{\cos\left(x\right)^{2}}}{x}}{\tan\left(x\right)^{2}}.

f(x) =
ddx[1xtan⁡(x)]=−1x2tan⁡(x)−1cos⁡(x)2xtan⁡(x)2\frac{d}{dx}\left[\frac{\frac{1}{x}}{\tan\left(x\right)}\right] = \frac{\frac{-1}{x^{2}} \tan\left(x\right) - \frac{\frac{1}{\cos\left(x\right)^{2}}}{x}}{\tan\left(x\right)^{2}}

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Graph of f(x) = 1 / x / tan(x) and its derivative-3-2-1123-2020406080slope -2.054f(x)f'(x)tangent at x = 1
Graph of f(x) = 1 / x / tan(x) and its derivative. The dashed line is the tangent at x = 1; its slope is f'(1) = -2.0544.

How to work it out

  1. 1

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

    f(x)=1xtan⁡(x)f(x) = \frac{\frac{1}{x}}{\tan\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=tan⁡(x)v = \tan\left(x\right).

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

    Differentiate each piece:

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

    Put them in the rule:

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

    Simplify:

    f′(x)=−1x2tan⁡(x)−1cos⁡(x)2xtan⁡(x)2f'(x) = \frac{\frac{-1}{x^{2}} \tan\left(x\right) - \frac{\frac{1}{\cos\left(x\right)^{2}}}{x}}{\tan\left(x\right)^{2}}
  6. 6

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

    f′′(x)=(−((−2)x)(x2)2tan⁡(x)+−1x21cos⁡(x)2−−(2cos⁡(x)−sin⁡(x))(cos⁡(x)2)2x−1cos⁡(x)2x2)tan⁡(x)2−(−1x2tan⁡(x)−1cos⁡(x)2x)2tan⁡(x)1cos⁡(x)2(tan⁡(x)2)2f''(x) = \frac{\left(\frac{-\left(\left(-2\right) x\right)}{\left(x^{2}\right)^{2}} \tan\left(x\right) + \frac{-1}{x^{2}} \frac{1}{\cos\left(x\right)^{2}} - \frac{\frac{-\left(2 \cos\left(x\right) -\sin\left(x\right)\right)}{\left(\cos\left(x\right)^{2}\right)^{2}} x - \frac{1}{\cos\left(x\right)^{2}}}{x^{2}}\right) \tan\left(x\right)^{2} - \left(\frac{-1}{x^{2}} \tan\left(x\right) - \frac{\frac{1}{\cos\left(x\right)^{2}}}{x}\right) 2 \tan\left(x\right) \frac{1}{\cos\left(x\right)^{2}}}{\left(\tan\left(x\right)^{2}\right)^{2}}
LaTeX source of the formulas on this page
\[ \frac{d}{dx}\left[\frac{\frac{1}{x}}{\tan\left(x\right)}\right] = \frac{\frac{-1}{x^{2}} \tan\left(x\right) - \frac{\frac{1}{\cos\left(x\right)^{2}}}{x}}{\tan\left(x\right)^{2}} \]
\[ f(x) = \frac{\frac{1}{x}}{\tan\left(x\right)} \]
\[ u = \frac{1}{x},\quad v = \tan\left(x\right) \]
\[ u' = \frac{-1}{x^{2}},\quad v' = \frac{1}{\cos\left(x\right)^{2}} \]
\[ f'(x) = \frac{\left(\frac{-1}{x^{2}}\right)\left(\tan\left(x\right)\right) - \left(\frac{1}{x}\right)\left(\frac{1}{\cos\left(x\right)^{2}}\right)}{\left(\tan\left(x\right)\right)^{2}} \]
\[ f'(x) = \frac{\frac{-1}{x^{2}} \tan\left(x\right) - \frac{\frac{1}{\cos\left(x\right)^{2}}}{x}}{\tan\left(x\right)^{2}} \]
\[ f''(x) = \frac{\left(\frac{-\left(\left(-2\right) x\right)}{\left(x^{2}\right)^{2}} \tan\left(x\right) + \frac{-1}{x^{2}} \frac{1}{\cos\left(x\right)^{2}} - \frac{\frac{-\left(2 \cos\left(x\right) -\sin\left(x\right)\right)}{\left(\cos\left(x\right)^{2}\right)^{2}} x - \frac{1}{\cos\left(x\right)^{2}}}{x^{2}}\right) \tan\left(x\right)^{2} - \left(\frac{-1}{x^{2}} \tan\left(x\right) - \frac{\frac{1}{\cos\left(x\right)^{2}}}{x}\right) 2 \tan\left(x\right) \frac{1}{\cos\left(x\right)^{2}}}{\left(\tan\left(x\right)^{2}\right)^{2}} \]

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

xf(x)f'(x)
-2-0.228828780.49031083
-10.642092622.0543755
0.53.6609754-16.023321
10.64209262-2.0543755
2-0.22882878-0.49031083
3-2.3384175-15.95845

Questions

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

The derivative of 1xtan⁡(x)\frac{\frac{1}{x}}{\tan\left(x\right)} is −1x2tan⁡(x)−1cos⁡(x)2xtan⁡(x)2\frac{\frac{-1}{x^{2}} \tan\left(x\right) - \frac{\frac{1}{\cos\left(x\right)^{2}}}{x}}{\tan\left(x\right)^{2}}.

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

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

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

The quotient rule.

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

-2.0543755.

Related derivatives

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