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Derivative of exx4\frac{e^{x}}{x^{4}}

The derivative of exx4\frac{e^{x}}{x^{4}} is exx4−ex4x3(x4)2\frac{e^{x} x^{4} - e^{x} 4 x^{3}}{\left(x^{4}\right)^{2}}.

f(x) =
ddx[exx4]=exx4−ex4x3(x4)2\frac{d}{dx}\left[\frac{e^{x}}{x^{4}}\right] = \frac{e^{x} x^{4} - e^{x} 4 x^{3}}{\left(x^{4}\right)^{2}}

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Graph of f(x) = exp(x) / x^4 and its derivative-3-2-11235001000150020002500slope -8.155f(x)f'(x)tangent at x = 1
Graph of f(x) = exp(x) / x^4 and its derivative. The dashed line is the tangent at x = 1; its slope is f'(1) = -8.1548.

How to work it out

  1. 1

    We differentiate f(x)=exx4f(x) = \frac{e^{x}}{x^{4}} with respect to x. The rule that applies is the quotient rule.

    f(x)=exx4f(x) = \frac{e^{x}}{x^{4}}
  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=x4v = x^{4}.

    u=ex,v=x4u = e^{x},\quad v = x^{4}
  3. 3

    Differentiate each piece:

    u′=ex,v′=4x3u' = e^{x},\quad v' = 4 x^{3}
  4. 4

    Put them in the rule:

    f′(x)=(ex)(x4)−(ex)(4x3)(x4)2f'(x) = \frac{\left(e^{x}\right)\left(x^{4}\right) - \left(e^{x}\right)\left(4 x^{3}\right)}{\left(x^{4}\right)^{2}}
  5. 5

    Simplify:

    f′(x)=exx4−ex4x3(x4)2f'(x) = \frac{e^{x} x^{4} - e^{x} 4 x^{3}}{\left(x^{4}\right)^{2}}
  6. 6

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

    f′′(x)=(exx4+ex4x3−(ex4x3+ex12x2))(x4)2−(exx4−ex4x3)2x44x3((x4)2)2f''(x) = \frac{\left(e^{x} x^{4} + e^{x} 4 x^{3} - \left(e^{x} 4 x^{3} + e^{x} 12 x^{2}\right)\right) \left(x^{4}\right)^{2} - \left(e^{x} x^{4} - e^{x} 4 x^{3}\right) 2 x^{4} 4 x^{3}}{\left(\left(x^{4}\right)^{2}\right)^{2}}
LaTeX source of the formulas on this page
\[ \frac{d}{dx}\left[\frac{e^{x}}{x^{4}}\right] = \frac{e^{x} x^{4} - e^{x} 4 x^{3}}{\left(x^{4}\right)^{2}} \]
\[ f(x) = \frac{e^{x}}{x^{4}} \]
\[ u = e^{x},\quad v = x^{4} \]
\[ u' = e^{x},\quad v' = 4 x^{3} \]
\[ f'(x) = \frac{\left(e^{x}\right)\left(x^{4}\right) - \left(e^{x}\right)\left(4 x^{3}\right)}{\left(x^{4}\right)^{2}} \]
\[ f'(x) = \frac{e^{x} x^{4} - e^{x} 4 x^{3}}{\left(x^{4}\right)^{2}} \]
\[ f''(x) = \frac{\left(e^{x} x^{4} + e^{x} 4 x^{3} - \left(e^{x} 4 x^{3} + e^{x} 12 x^{2}\right)\right) \left(x^{4}\right)^{2} - \left(e^{x} x^{4} - e^{x} 4 x^{3}\right) 2 x^{4} 4 x^{3}}{\left(\left(x^{4}\right)^{2}\right)^{2}} \]

Values of f and f' for exp(x) / x4x^{4}

xf(x)f'(x)
-20.00845845520.025375366
-10.367879441.8393972
0.526.37954-184.65678
12.7182818-8.1548455
20.46181601-0.46181601
30.24796959-0.082656531

Questions

What is the derivative of exp(x) / x4x^{4}?

The derivative of exx4\frac{e^{x}}{x^{4}} is exx4−ex4x3(x4)2\frac{e^{x} x^{4} - e^{x} 4 x^{3}}{\left(x^{4}\right)^{2}}.

What is the second derivative of exp(x) / x4x^{4}?

((exp(x) * x4x^{4} + exp(x) * 4 * x3x^{3} - (exp(x) * 4 * x3x^{3} + exp(x) * 12 * x2x^{2})) * (x4x^{4})^2 - (exp(x) * x4x^{4} - exp(x) * 4 * x3x^{3}) * 2 * x4x^{4} * 4 * x3x^{3}) / ((x4x^{4})^2)^2.

Which rule do you use for exp(x) / x4x^{4}?

The quotient rule.

What is the slope of exp(x) / x4x^{4} at x = 1?

-8.1548455.

Related derivatives

    Derivative of exp(x) / x^4: (exp(x) * x^4 - exp(x) * 4 * x^3) / (x^4)^2 (with steps) | WordGPT