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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
For x = 1
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.
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