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

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

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

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Graph of f(x) = 1 / x * tan(x) and its derivative-3-2-1123-15-10-551015slope 1.868f(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) = 1.8681.

How to work it out

  1. 1

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

    f(x)=1xtan⁡(x)f(x) = \frac{1}{x} \tan\left(x\right)
  2. 2

    Product rule: (uv)′=u′v+uv′(uv)' = u'v + uv'. 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)f'(x) = \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)
  5. 5

    Simplify:

    f′(x)=−1x2tan⁡(x)+1cos⁡(x)2xf'(x) = \frac{-1}{x^{2}} \tan\left(x\right) + \frac{\frac{1}{\cos\left(x\right)^{2}}}{x}
  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)2x2f''(x) = \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}}
LaTeX source of the formulas on this page
\[ \frac{d}{dx}\left[\frac{1}{x} \tan\left(x\right)\right] = \frac{-1}{x^{2}} \tan\left(x\right) + \frac{\frac{1}{\cos\left(x\right)^{2}}}{x} \]
\[ f(x) = \frac{1}{x} \tan\left(x\right) \]
\[ u = \frac{1}{x},\quad v = \tan\left(x\right) \]
\[ u' = \frac{-1}{x^{2}},\quad v' = \frac{1}{\cos\left(x\right)^{2}} \]
\[ f'(x) = \left(\frac{-1}{x^{2}}\right)\left(\tan\left(x\right)\right) + \left(\frac{1}{x}\right)\left(\frac{1}{\cos\left(x\right)^{2}}\right) \]
\[ f'(x) = \frac{-1}{x^{2}} \tan\left(x\right) + \frac{\frac{1}{\cos\left(x\right)^{2}}}{x} \]
\[ f''(x) = \frac{-\left(\left(-2\right) x\right)}{\left(x^{2}\right)^{2}} \tan\left(x\right) + \frac{-1}{x^{2}} \frac{1}{\cos\left(x\right)^{2}} + \frac{\frac{-\left(2 \cos\left(x\right) -\sin\left(x\right)\right)}{\left(\cos\left(x\right)^{2}\right)^{2}} x - \frac{1}{\cos\left(x\right)^{2}}}{x^{2}} \]

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

xf(x)f'(x)
-2-1.0925199-3.4334596
-11.5574077-1.8681111
0.51.0926050.41168286
11.55740771.8681111
2-1.09251993.4334596
3-0.0475155140.35594501

Questions

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

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

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}.

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

The product rule.

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

1.8681111.

Related derivatives

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