# Find the slope of the tangent to the curve f(x) = e^(-x^2) at the point on the curve where x = 1.

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### 2 Answers

We need the slope of the tangent to the curve f(x) = e^(-x^2) at the point on the curve where x = 1.

This is the value of the first derivative of f(x) at x = 1.

f'(x) = [e^(-x^2)]'

use the chain rule

=> e^(-x^2)*(-2x)

For x = 1

=> (e^-1)*(-2)

=> -2/e

**The required slope of the tangent to the slope f(x) = e^(-x^2) at the point x = 1 is -2/e.**

First find the slope at x = 1.

We know the derivative is the instantanious slope so

f '(x) = e^(-x^2) * -2x = -2x e^(-x^2)

f '(1) = -2(1)e^(-(1)^2) = -2 e^(-1) = -2/e

**So our answer is -2/e is the slope of the tangent to the curve.**