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Derivative of cos⁡(x)x3\frac{\cos\left(x\right)}{x^{3}}

The derivative of cos⁡(x)x3\frac{\cos\left(x\right)}{x^{3}} is −(sin⁡(x)x3)−cos⁡(x)3x2(x3)2\frac{-\left(\sin\left(x\right) x^{3}\right) - \cos\left(x\right) 3 x^{2}}{\left(x^{3}\right)^{2}}.

f(x) =
ddx[cos⁡(x)x3]=−(sin⁡(x)x3)−cos⁡(x)3x2(x3)2\frac{d}{dx}\left[\frac{\cos\left(x\right)}{x^{3}}\right] = \frac{-\left(\sin\left(x\right) x^{3}\right) - \cos\left(x\right) 3 x^{2}}{\left(x^{3}\right)^{2}}

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Graph of f(x) = cos(x) / x^3 and its derivative-3-2-1123-1000-800-600-400-200slope -2.462f(x)f'(x)tangent at x = 1
Graph of f(x) = cos(x) / x^3 and its derivative. The dashed line is the tangent at x = 1; its slope is f'(1) = -2.4624.

How to work it out

  1. 1

    We differentiate f(x)=cos⁡(x)x3f(x) = \frac{\cos\left(x\right)}{x^{3}} with respect to x. The rule that applies is the quotient rule.

    f(x)=cos⁡(x)x3f(x) = \frac{\cos\left(x\right)}{x^{3}}
  2. 2

    Quotient rule: (uv)′=u′v−uv′v2\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^{2}}. Let u=cos⁡(x)u = \cos\left(x\right) and v=x3v = x^{3}.

    u=cos⁡(x),v=x3u = \cos\left(x\right),\quad v = x^{3}
  3. 3

    Differentiate each piece:

    u′=−sin⁡(x),v′=3x2u' = -\sin\left(x\right),\quad v' = 3 x^{2}
  4. 4

    Put them in the rule:

    f′(x)=(−sin⁡(x))(x3)−(cos⁡(x))(3x2)(x3)2f'(x) = \frac{\left(-\sin\left(x\right)\right)\left(x^{3}\right) - \left(\cos\left(x\right)\right)\left(3 x^{2}\right)}{\left(x^{3}\right)^{2}}
  5. 5

    Simplify:

    f′(x)=−(sin⁡(x)x3)−cos⁡(x)3x2(x3)2f'(x) = \frac{-\left(\sin\left(x\right) x^{3}\right) - \cos\left(x\right) 3 x^{2}}{\left(x^{3}\right)^{2}}
  6. 6

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

    f′′(x)=(−(cos⁡(x)x3+sin⁡(x)3x2)−(−(sin⁡(x)3x2)+cos⁡(x)6x))(x3)2−(−(sin⁡(x)x3)−cos⁡(x)3x2)2x33x2((x3)2)2f''(x) = \frac{\left(-\left(\cos\left(x\right) x^{3} + \sin\left(x\right) 3 x^{2}\right) - \left(-\left(\sin\left(x\right) 3 x^{2}\right) + \cos\left(x\right) 6 x\right)\right) \left(x^{3}\right)^{2} - \left(-\left(\sin\left(x\right) x^{3}\right) - \cos\left(x\right) 3 x^{2}\right) 2 x^{3} 3 x^{2}}{\left(\left(x^{3}\right)^{2}\right)^{2}}
LaTeX source of the formulas on this page
\[ \frac{d}{dx}\left[\frac{\cos\left(x\right)}{x^{3}}\right] = \frac{-\left(\sin\left(x\right) x^{3}\right) - \cos\left(x\right) 3 x^{2}}{\left(x^{3}\right)^{2}} \]
\[ f(x) = \frac{\cos\left(x\right)}{x^{3}} \]
\[ u = \cos\left(x\right),\quad v = x^{3} \]
\[ u' = -\sin\left(x\right),\quad v' = 3 x^{2} \]
\[ f'(x) = \frac{\left(-\sin\left(x\right)\right)\left(x^{3}\right) - \left(\cos\left(x\right)\right)\left(3 x^{2}\right)}{\left(x^{3}\right)^{2}} \]
\[ f'(x) = \frac{-\left(\sin\left(x\right) x^{3}\right) - \cos\left(x\right) 3 x^{2}}{\left(x^{3}\right)^{2}} \]
\[ f''(x) = \frac{\left(-\left(\cos\left(x\right) x^{3} + \sin\left(x\right) 3 x^{2}\right) - \left(-\left(\sin\left(x\right) 3 x^{2}\right) + \cos\left(x\right) 6 x\right)\right) \left(x^{3}\right)^{2} - \left(-\left(\sin\left(x\right) x^{3}\right) - \cos\left(x\right) 3 x^{2}\right) 2 x^{3} 3 x^{2}}{\left(\left(x^{3}\right)^{2}\right)^{2}} \]

Values of f and f' for cos(x) / x3x^{3}

xf(x)f'(x)
-20.052018355-0.035634647
-1-0.54030231-2.4623779
0.57.0206605-45.959367
10.54030231-2.4623779
2-0.052018355-0.035634647
3-0.0366663890.031439722

Questions

What is the derivative of cos(x) / x3x^{3}?

The derivative of cos⁡(x)x3\frac{\cos\left(x\right)}{x^{3}} is −(sin⁡(x)x3)−cos⁡(x)3x2(x3)2\frac{-\left(\sin\left(x\right) x^{3}\right) - \cos\left(x\right) 3 x^{2}}{\left(x^{3}\right)^{2}}.

What is the second derivative of cos(x) / x3x^{3}?

((-(cos(x) * x3x^{3} + sin(x) * 3 * x2x^{2}) - (-(sin(x) * 3 * x2x^{2}) + cos(x) * 6x)) * (x3x^{3})^2 - (-(sin(x) * x3x^{3}) - cos(x) * 3 * x2x^{2}) * 2 * x3x^{3} * 3 * x2x^{2}) / ((x3x^{3})^2)^2.

Which rule do you use for cos(x) / x3x^{3}?

The quotient rule.

What is the slope of cos(x) / x3x^{3} at x = 1?

-2.4623779.

Related derivatives

    Derivative of cos(x) / x^3: (-(sin(x) * x^3) - cos(x) * 3 * x^2) / (x^3)^2 (with steps) | WordGPT