Find an equation of the tangent line to the curve at the point (0,0).

y=x

Step-by-step 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

step-by-step explanation:

1/8 3/16 3/4

step-by-step explanation:

you put them as all the same denominator and multiply