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Derivative of exx5\frac{e^{x}}{x^{5}}

The derivative of exx5\frac{e^{x}}{x^{5}} is exx5−ex5x4(x5)2\frac{e^{x} x^{5} - e^{x} 5 x^{4}}{\left(x^{5}\right)^{2}}.

f(x) =
ddx[exx5]=exx5−ex5x4(x5)2\frac{d}{dx}\left[\frac{e^{x}}{x^{5}}\right] = \frac{e^{x} x^{5} - e^{x} 5 x^{4}}{\left(x^{5}\right)^{2}}

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Graph of f(x) = exp(x) / x^5 and its derivative-3-2-1123-40000-30000-20000-10000slope -10.87f(x)f'(x)tangent at x = 1
Graph of f(x) = exp(x) / x^5 and its derivative. The dashed line is the tangent at x = 1; its slope is f'(1) = -10.873.

How to work it out

  1. 1

    We differentiate f(x)=exx5f(x) = \frac{e^{x}}{x^{5}} with respect to x. The rule that applies is the quotient rule.

    f(x)=exx5f(x) = \frac{e^{x}}{x^{5}}
  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=x5v = x^{5}.

    u=ex,v=x5u = e^{x},\quad v = x^{5}
  3. 3

    Differentiate each piece:

    u′=ex,v′=5x4u' = e^{x},\quad v' = 5 x^{4}
  4. 4

    Put them in the rule:

    f′(x)=(ex)(x5)−(ex)(5x4)(x5)2f'(x) = \frac{\left(e^{x}\right)\left(x^{5}\right) - \left(e^{x}\right)\left(5 x^{4}\right)}{\left(x^{5}\right)^{2}}
  5. 5

    Simplify:

    f′(x)=exx5−ex5x4(x5)2f'(x) = \frac{e^{x} x^{5} - e^{x} 5 x^{4}}{\left(x^{5}\right)^{2}}
  6. 6

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

    f′′(x)=(exx5+ex5x4−(ex5x4+ex20x3))(x5)2−(exx5−ex5x4)2x55x4((x5)2)2f''(x) = \frac{\left(e^{x} x^{5} + e^{x} 5 x^{4} - \left(e^{x} 5 x^{4} + e^{x} 20 x^{3}\right)\right) \left(x^{5}\right)^{2} - \left(e^{x} x^{5} - e^{x} 5 x^{4}\right) 2 x^{5} 5 x^{4}}{\left(\left(x^{5}\right)^{2}\right)^{2}}
LaTeX source of the formulas on this page
\[ \frac{d}{dx}\left[\frac{e^{x}}{x^{5}}\right] = \frac{e^{x} x^{5} - e^{x} 5 x^{4}}{\left(x^{5}\right)^{2}} \]
\[ f(x) = \frac{e^{x}}{x^{5}} \]
\[ u = e^{x},\quad v = x^{5} \]
\[ u' = e^{x},\quad v' = 5 x^{4} \]
\[ f'(x) = \frac{\left(e^{x}\right)\left(x^{5}\right) - \left(e^{x}\right)\left(5 x^{4}\right)}{\left(x^{5}\right)^{2}} \]
\[ f'(x) = \frac{e^{x} x^{5} - e^{x} 5 x^{4}}{\left(x^{5}\right)^{2}} \]
\[ f''(x) = \frac{\left(e^{x} x^{5} + e^{x} 5 x^{4} - \left(e^{x} 5 x^{4} + e^{x} 20 x^{3}\right)\right) \left(x^{5}\right)^{2} - \left(e^{x} x^{5} - e^{x} 5 x^{4}\right) 2 x^{5} 5 x^{4}}{\left(\left(x^{5}\right)^{2}\right)^{2}} \]

Values of f and f' for exp(x) / x5x^{5}

xf(x)f'(x)
-2-0.0042292276-0.014802297
-1-0.36787944-2.2072766
0.552.759081-474.83173
12.7182818-10.873127
20.230908-0.346362
30.082656531-0.055104354

Questions

What is the derivative of exp(x) / x5x^{5}?

The derivative of exx5\frac{e^{x}}{x^{5}} is exx5−ex5x4(x5)2\frac{e^{x} x^{5} - e^{x} 5 x^{4}}{\left(x^{5}\right)^{2}}.

What is the second derivative of exp(x) / x5x^{5}?

((exp(x) * x5x^{5} + exp(x) * 5 * x4x^{4} - (exp(x) * 5 * x4x^{4} + exp(x) * 20 * x3x^{3})) * (x5x^{5})^2 - (exp(x) * x5x^{5} - exp(x) * 5 * x4x^{4}) * 2 * x5x^{5} * 5 * x4x^{4}) / ((x5x^{5})^2)^2.

Which rule do you use for exp(x) / x5x^{5}?

The quotient rule.

What is the slope of exp(x) / x5x^{5} at x = 1?

-10.873127.

Related derivatives

    Derivative of exp(x) / x^5: (exp(x) * x^5 - exp(x) * 5 * x^4) / (x^5)^2 (with steps) | WordGPT