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Derivative of 1xcos⁡(x)\frac{\frac{1}{x}}{\cos\left(x\right)}

The derivative of 1xcos⁡(x)\frac{\frac{1}{x}}{\cos\left(x\right)} is −1x2cos⁡(x)−−sin⁡(x)xcos⁡(x)2\frac{\frac{-1}{x^{2}} \cos\left(x\right) - \frac{-\sin\left(x\right)}{x}}{\cos\left(x\right)^{2}}.

f(x) =
ddx[1xcos⁡(x)]=−1x2cos⁡(x)−−sin⁡(x)xcos⁡(x)2\frac{d}{dx}\left[\frac{\frac{1}{x}}{\cos\left(x\right)}\right] = \frac{\frac{-1}{x^{2}} \cos\left(x\right) - \frac{-\sin\left(x\right)}{x}}{\cos\left(x\right)^{2}}

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Graph of f(x) = 1 / x / cos(x) and its derivative-3-2-1123-20-1010203040slope 1.032f(x)f'(x)tangent at x = 1
Graph of f(x) = 1 / x / cos(x) and its derivative. The dashed line is the tangent at x = 1; its slope is f'(1) = 1.0317.

How to work it out

  1. 1

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

    f(x)=1xcos⁡(x)f(x) = \frac{\frac{1}{x}}{\cos\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=cos⁡(x)v = \cos\left(x\right).

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

    Differentiate each piece:

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

    Put them in the rule:

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

    Simplify:

    f′(x)=−1x2cos⁡(x)−−sin⁡(x)xcos⁡(x)2f'(x) = \frac{\frac{-1}{x^{2}} \cos\left(x\right) - \frac{-\sin\left(x\right)}{x}}{\cos\left(x\right)^{2}}
  6. 6

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

    f′′(x)=(−((−2)x)(x2)2cos⁡(x)+−1x2−sin⁡(x)−−(cos⁡(x)x)−−sin⁡(x)x2)cos⁡(x)2−(−1x2cos⁡(x)−−sin⁡(x)x)2cos⁡(x)−sin⁡(x)(cos⁡(x)2)2f''(x) = \frac{\left(\frac{-\left(\left(-2\right) x\right)}{\left(x^{2}\right)^{2}} \cos\left(x\right) + \frac{-1}{x^{2}} -\sin\left(x\right) - \frac{-\left(\cos\left(x\right) x\right) - -\sin\left(x\right)}{x^{2}}\right) \cos\left(x\right)^{2} - \left(\frac{-1}{x^{2}} \cos\left(x\right) - \frac{-\sin\left(x\right)}{x}\right) 2 \cos\left(x\right) -\sin\left(x\right)}{\left(\cos\left(x\right)^{2}\right)^{2}}
LaTeX source of the formulas on this page
\[ \frac{d}{dx}\left[\frac{\frac{1}{x}}{\cos\left(x\right)}\right] = \frac{\frac{-1}{x^{2}} \cos\left(x\right) - \frac{-\sin\left(x\right)}{x}}{\cos\left(x\right)^{2}} \]
\[ f(x) = \frac{\frac{1}{x}}{\cos\left(x\right)} \]
\[ u = \frac{1}{x},\quad v = \cos\left(x\right) \]
\[ u' = \frac{-1}{x^{2}},\quad v' = -\sin\left(x\right) \]
\[ f'(x) = \frac{\left(\frac{-1}{x^{2}}\right)\left(\cos\left(x\right)\right) - \left(\frac{1}{x}\right)\left(-\sin\left(x\right)\right)}{\left(\cos\left(x\right)\right)^{2}} \]
\[ f'(x) = \frac{\frac{-1}{x^{2}} \cos\left(x\right) - \frac{-\sin\left(x\right)}{x}}{\cos\left(x\right)^{2}} \]
\[ f''(x) = \frac{\left(\frac{-\left(\left(-2\right) x\right)}{\left(x^{2}\right)^{2}} \cos\left(x\right) + \frac{-1}{x^{2}} -\sin\left(x\right) - \frac{-\left(\cos\left(x\right) x\right) - -\sin\left(x\right)}{x^{2}}\right) \cos\left(x\right)^{2} - \left(\frac{-1}{x^{2}} \cos\left(x\right) - \frac{-\sin\left(x\right)}{x}\right) 2 \cos\left(x\right) -\sin\left(x\right)}{\left(\cos\left(x\right)^{2}\right)^{2}} \]

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

xf(x)f'(x)
-21.2014993.2260727
-1-1.85081571.031659
0.52.2789879-3.312959
11.85081571.031659
2-1.2014993.2260727
3-0.336702890.16023013

Questions

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

The derivative of 1xcos⁡(x)\frac{\frac{1}{x}}{\cos\left(x\right)} is −1x2cos⁡(x)−−sin⁡(x)xcos⁡(x)2\frac{\frac{-1}{x^{2}} \cos\left(x\right) - \frac{-\sin\left(x\right)}{x}}{\cos\left(x\right)^{2}}.

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

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

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

The quotient rule.

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

1.031659.

Related derivatives

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