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Derivative of ex2e^{x^{2}}

The derivative of ex2e^{x^{2}} is ex22xe^{x^{2}} 2 x.

f(x) =
ddx[ex2]=ex22x\frac{d}{dx}\left[e^{x^{2}}\right] = e^{x^{2}} 2 x

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Graph of f(x) = exp(x^2) and its derivative-3-2-1123-2000200040006000slope 5.437f(x)f'(x)tangent at x = 1
Graph of f(x) = exp(x^2) 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)=ex2f(x) = e^{x^{2}} with respect to x. The rule that applies is the chain rule.

    f(x)=ex2f(x) = e^{x^{2}}
  2. 2

    Chain rule: differentiate the outer function, keep the inside, then multiply by the derivative of the inside. The inside is g(x)=x2g(x) = x^{2}.

    g(x)=x2,g′(x)=2xg(x) = x^{2},\quad g'(x) = 2 x
  3. 3

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

    f′(x)=[eu]u=g(x)⋅g′(x)f'(x) = \left[e^{u}\right]_{u=g(x)} \cdot g'(x)
  4. 4

    Simplify:

    f′(x)=ex22xf'(x) = e^{x^{2}} 2 x
  5. 5

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

    f′′(x)=ex22x2x+2ex2f''(x) = e^{x^{2}} 2 x 2 x + 2 e^{x^{2}}
LaTeX source of the formulas on this page
\[ \frac{d}{dx}\left[e^{x^{2}}\right] = e^{x^{2}} 2 x \]
\[ f(x) = e^{x^{2}} \]
\[ g(x) = x^{2},\quad g'(x) = 2 x \]
\[ f'(x) = \left[e^{u}\right]_{u=g(x)} \cdot g'(x) \]
\[ f'(x) = e^{x^{2}} 2 x \]
\[ f''(x) = e^{x^{2}} 2 x 2 x + 2 e^{x^{2}} \]

Values of f and f' for exp(x2x^{2})

xf(x)f'(x)
-254.59815-218.3926
-12.7182818-5.4365637
0.51.28402541.2840254
12.71828185.4365637
254.59815218.3926
38103.083948618.504

Questions

What is the derivative of exp(x2x^{2})?

The derivative of ex2e^{x^{2}} is ex22xe^{x^{2}} 2 x.

What is the second derivative of exp(x2x^{2})?

exp(x2x^{2}) * 2x * 2x + 2exp(x2x^{2}).

Which rule do you use for exp(x2x^{2})?

The chain rule.

What is the slope of exp(x2x^{2}) at x = 1?

5.4365637.

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