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Derivative of exx3\frac{e^{x}}{x^{3}}

The derivative of exx3\frac{e^{x}}{x^{3}} is exx3−ex3x2(x3)2\frac{e^{x} x^{3} - e^{x} 3 x^{2}}{\left(x^{3}\right)^{2}}.

f(x) =
ddx[exx3]=exx3−ex3x2(x3)2\frac{d}{dx}\left[\frac{e^{x}}{x^{3}}\right] = \frac{e^{x} x^{3} - e^{x} 3 x^{2}}{\left(x^{3}\right)^{2}}

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Graph of f(x) = exp(x) / x^3 and its derivative-3-2-1123-1000-800-600-400-200slope -5.437f(x)f'(x)tangent at x = 1
Graph of f(x) = exp(x) / x^3 and its derivative. The dashed line is the tangent at x = 1; its slope is f'(1) = -5.4366.

How to work it out

  1. 1

    We differentiate f(x)=exx3f(x) = \frac{e^{x}}{x^{3}} with respect to x. The rule that applies is the quotient rule.

    f(x)=exx3f(x) = \frac{e^{x}}{x^{3}}
  2. 2

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

    u=ex,v=x3u = e^{x},\quad v = x^{3}
  3. 3

    Differentiate each piece:

    u′=ex,v′=3x2u' = e^{x},\quad v' = 3 x^{2}
  4. 4

    Put them in the rule:

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

    Simplify:

    f′(x)=exx3−ex3x2(x3)2f'(x) = \frac{e^{x} x^{3} - e^{x} 3 x^{2}}{\left(x^{3}\right)^{2}}
  6. 6

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

    f′′(x)=(exx3+ex3x2−(ex3x2+ex6x))(x3)2−(exx3−ex3x2)2x33x2((x3)2)2f''(x) = \frac{\left(e^{x} x^{3} + e^{x} 3 x^{2} - \left(e^{x} 3 x^{2} + e^{x} 6 x\right)\right) \left(x^{3}\right)^{2} - \left(e^{x} x^{3} - e^{x} 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
\[ \frac{d}{dx}\left[\frac{e^{x}}{x^{3}}\right] = \frac{e^{x} x^{3} - e^{x} 3 x^{2}}{\left(x^{3}\right)^{2}} \]
\[ f(x) = \frac{e^{x}}{x^{3}} \]
\[ u = e^{x},\quad v = x^{3} \]
\[ u' = e^{x},\quad v' = 3 x^{2} \]
\[ f'(x) = \frac{\left(e^{x}\right)\left(x^{3}\right) - \left(e^{x}\right)\left(3 x^{2}\right)}{\left(x^{3}\right)^{2}} \]
\[ f'(x) = \frac{e^{x} x^{3} - e^{x} 3 x^{2}}{\left(x^{3}\right)^{2}} \]
\[ f''(x) = \frac{\left(e^{x} x^{3} + e^{x} 3 x^{2} - \left(e^{x} 3 x^{2} + e^{x} 6 x\right)\right) \left(x^{3}\right)^{2} - \left(e^{x} x^{3} - e^{x} 3 x^{2}\right) 2 x^{3} 3 x^{2}}{\left(\left(x^{3}\right)^{2}\right)^{2}} \]

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

xf(x)f'(x)
-2-0.01691691-0.042292276
-1-0.36787944-1.4715178
0.513.18977-65.948851
12.7182818-5.4365637
20.92363201-0.46181601
30.743908770

Questions

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

The derivative of exx3\frac{e^{x}}{x^{3}} is exx3−ex3x2(x3)2\frac{e^{x} x^{3} - e^{x} 3 x^{2}}{\left(x^{3}\right)^{2}}.

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

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

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

The quotient rule.

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

-5.4365637.

Related derivatives

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