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Derivative of e1āˆ’xe^{1 - x}

The derivative of e1āˆ’xe^{1 - x} is āˆ’e1āˆ’x-e^{1 - x}.

f(x) =
ddx[e1āˆ’x]=āˆ’e1āˆ’x\frac{d}{dx}\left[e^{1 - x}\right] = -e^{1 - x}

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Graph of f(x) = exp(1 - x) and its derivative-3-2-1123-40-30-20-1010203040slope -1f(x)f'(x)tangent at x = 1
Graph of f(x) = exp(1 - x) and its derivative. The dashed line is the tangent at x = 1; its slope is f'(1) = -1.

How to work it out

  1. 1

    We differentiate f(x)=e1āˆ’xf(x) = e^{1 - x} with respect to x. The rule that applies is the chain rule.

    f(x)=e1āˆ’xf(x) = e^{1 - 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)=1āˆ’xg(x) = 1 - x.

    g(x)=1āˆ’x,gā€²(x)=āˆ’1g(x) = 1 - 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’e1āˆ’xf'(x) = -e^{1 - x}
  5. 5

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

    fā€²ā€²(x)=e1āˆ’xf''(x) = e^{1 - x}
LaTeX source of the formulas on this page
\[ \frac{d}{dx}\left[e^{1 - x}\right] = -e^{1 - x} \]
\[ f(x) = e^{1 - x} \]
\[ g(x) = 1 - x,\quad g'(x) = -1 \]
\[ f'(x) = \left[e^{u}\right]_{u=g(x)} \cdot g'(x) \]
\[ f'(x) = -e^{1 - x} \]
\[ f''(x) = e^{1 - x} \]

Values of f and f' for exp(1 - x)

xf(x)f'(x)
-220.085537-20.085537
-17.3890561-7.3890561
0.51.6487213-1.6487213
11-1
20.36787944-0.36787944
30.13533528-0.13533528

Questions

What is the derivative of exp(1 - x)?

The derivative of e1āˆ’xe^{1 - x} is āˆ’e1āˆ’x-e^{1 - x}.

What is the second derivative of exp(1 - x)?

exp(1 - x).

Which rule do you use for exp(1 - x)?

The chain rule.

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

-1.

Related derivatives

    Derivative of exp(1 - x): -exp(1 - x) (with steps) | WordGPT