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

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

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

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Graph of f(x) = 1 / x * cos(x) and its derivative-3-2-1123-15-10-55slope -1.382f(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) = -1.3818.

How to work it out

  1. 1

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

    f(x)=1xcos⁡(x)f(x) = \frac{1}{x} \cos\left(x\right)
  2. 2

    Product rule: (uv)′=u′v+uv′(uv)' = u'v + uv'. 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))f'(x) = \left(\frac{-1}{x^{2}}\right)\left(\cos\left(x\right)\right) + \left(\frac{1}{x}\right)\left(-\sin\left(x\right)\right)
  5. 5

    Simplify:

    f′(x)=−1x2cos⁡(x)+−sin⁡(x)xf'(x) = \frac{-1}{x^{2}} \cos\left(x\right) + \frac{-\sin\left(x\right)}{x}
  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)x2f''(x) = \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}}
LaTeX source of the formulas on this page
\[ \frac{d}{dx}\left[\frac{1}{x} \cos\left(x\right)\right] = \frac{-1}{x^{2}} \cos\left(x\right) + \frac{-\sin\left(x\right)}{x} \]
\[ f(x) = \frac{1}{x} \cos\left(x\right) \]
\[ u = \frac{1}{x},\quad v = \cos\left(x\right) \]
\[ u' = \frac{-1}{x^{2}},\quad v' = -\sin\left(x\right) \]
\[ f'(x) = \left(\frac{-1}{x^{2}}\right)\left(\cos\left(x\right)\right) + \left(\frac{1}{x}\right)\left(-\sin\left(x\right)\right) \]
\[ f'(x) = \frac{-1}{x^{2}} \cos\left(x\right) + \frac{-\sin\left(x\right)}{x} \]
\[ f''(x) = \frac{-\left(\left(-2\right) x\right)}{\left(x^{2}\right)^{2}} \cos\left(x\right) + \frac{-1}{x^{2}} -\sin\left(x\right) + \frac{-\left(\cos\left(x\right) x\right) - -\sin\left(x\right)}{x^{2}} \]

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

xf(x)f'(x)
-20.20807342-0.350612
-1-0.54030231-1.3817733
0.51.7551651-4.4691813
10.54030231-1.3817733
2-0.20807342-0.350612
3-0.32999750.062959164

Questions

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

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

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}.

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

The product rule.

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

-1.3817733.

Related derivatives

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