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Find an equation of the tangent line to the curve
y=xe^{-x^(2) } at the point (0,0).

1 Answer

5 votes

Answer:

y=x

Explanation:

We must find the derivative (the formula for the slopes of the tangent lines at any x).

We will need product rule:

y'=(1)e^(-x^2)+x(-2x)e^(-x^2)

y'=e^(-x^2)-2x^2e^(-x^2)

At x=0 we get the slope is e^(-0^2)-2(0)^2e^(-0^2)

=1-0

=1.

So the line going through (0,0) with slope 1 is

y-0=1(x-0)

Or

y=x

User Ammar Mujeeb
by
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