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

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

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

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Graph of f(x) = 1 / x * sin(x) and its derivative-3-2-1123-0.50.511.52slope -0.3012f(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) = -0.30117.

How to work it out

  1. 1

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

    f(x)=1xsin⁡(x)f(x) = \frac{1}{x} \sin\left(x\right)
  2. 2

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

    Simplify:

    f′(x)=−1x2sin⁡(x)+cos⁡(x)xf'(x) = \frac{-1}{x^{2}} \sin\left(x\right) + \frac{\cos\left(x\right)}{x}
  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)x2f''(x) = \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}}
LaTeX source of the formulas on this page
\[ \frac{d}{dx}\left[\frac{1}{x} \sin\left(x\right)\right] = \frac{-1}{x^{2}} \sin\left(x\right) + \frac{\cos\left(x\right)}{x} \]
\[ f(x) = \frac{1}{x} \sin\left(x\right) \]
\[ u = \frac{1}{x},\quad v = \sin\left(x\right) \]
\[ u' = \frac{-1}{x^{2}},\quad v' = \cos\left(x\right) \]
\[ f'(x) = \left(\frac{-1}{x^{2}}\right)\left(\sin\left(x\right)\right) + \left(\frac{1}{x}\right)\left(\cos\left(x\right)\right) \]
\[ f'(x) = \frac{-1}{x^{2}} \sin\left(x\right) + \frac{\cos\left(x\right)}{x} \]
\[ f''(x) = \frac{-\left(\left(-2\right) x\right)}{\left(x^{2}\right)^{2}} \sin\left(x\right) + \frac{-1}{x^{2}} \cos\left(x\right) + \frac{-\left(\sin\left(x\right) x\right) - \cos\left(x\right)}{x^{2}} \]

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

xf(x)f'(x)
-20.454648710.43539777
-10.841470980.30116868
0.50.95885108-0.16253703
10.84147098-0.30116868
20.45464871-0.43539777
30.047040003-0.3456775

Questions

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

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

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}.

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

The product rule.

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

-0.30116868.

Related derivatives

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