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Derivative of eāˆ’xe^{-x}

The derivative of eāˆ’xe^{-x} is āˆ’eāˆ’x-e^{-x}.

f(x) =
ddx[eāˆ’x]=āˆ’eāˆ’x\frac{d}{dx}\left[e^{-x}\right] = -e^{-x}

Type any function of x; the derivative is computed symbolically in your browser.

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Graph of f(x) = exp(-x) and its derivative-3-2-1123-15-10-551015slope -0.3679f(x)f'(x)tangent at x = 1
Graph of f(x) = exp(-x) and its derivative. The dashed line is the tangent at x = 1; its slope is f'(1) = -0.36788.

How to work it out

  1. 1

    We differentiate f(x)=eāˆ’xf(x) = e^{-x} with respect to x. The rule that applies is the chain rule.

    f(x)=eāˆ’xf(x) = e^{-x}
  2. 2

    Chain rule: differentiate the outer function, keep the inside, then multiply by the derivative of the inside. The inside is g(x)=āˆ’xg(x) = -x.

    g(x)=āˆ’x,gā€²(x)=āˆ’1g(x) = -x,\quad g'(x) = -1
  3. 3

    The outer function is eue^{u}; its derivative evaluated at g(x), times g'(x):

    fā€²(x)=[eu]u=g(x)ā‹…gā€²(x)f'(x) = \left[e^{u}\right]_{u=g(x)} \cdot g'(x)
  4. 4

    Simplify:

    fā€²(x)=āˆ’eāˆ’xf'(x) = -e^{-x}
  5. 5

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

    fā€²ā€²(x)=eāˆ’xf''(x) = e^{-x}
LaTeX source of the formulas on this page
\[ \frac{d}{dx}\left[e^{-x}\right] = -e^{-x} \]
\[ f(x) = e^{-x} \]
\[ g(x) = -x,\quad g'(x) = -1 \]
\[ f'(x) = \left[e^{u}\right]_{u=g(x)} \cdot g'(x) \]
\[ f'(x) = -e^{-x} \]
\[ f''(x) = e^{-x} \]

Values of f and f' for exp(-x)

xf(x)f'(x)
-27.3890561-7.3890561
-12.7182818-2.7182818
0.50.60653066-0.60653066
10.36787944-0.36787944
20.13533528-0.13533528
30.049787068-0.049787068

Questions

What is the derivative of exp(-x)?

The derivative of eāˆ’xe^{-x} is āˆ’eāˆ’x-e^{-x}.

What is the second derivative of exp(-x)?

exp(-x).

Which rule do you use for exp(-x)?

The chain rule.

What is the slope of exp(-x) at x = 1?

-0.36787944.

Related derivatives

    Derivative of exp(-x): -exp(-x) (with steps) | WordGPT