how to find the equation: y = x^2 * e^(-x) of the tangent line to points (1, 1/e)?

Answer 1

When asked to find the tangent line, you should immediately think of the derivative.
To find the derivative of y, you will need to use the product rule.
y = x^2 * e^(-x)
y' = (x^2)' *e^(-x) + x^2 (e^(-x))'
y' = (2x)*e^(-x) + x^2 (-e^(-x))
y' = 2x *e^(-x) - x^2 *e^(-x)
This will give you the slope of the tangent line at any point. The point in particular is (1,1/e). Replace x with 1 to find the slope.
y'(1) = 2(1)e^(-1) - 1^2 *e^(-1)
y'(1) = 2e^(-1) - e^(-1)
y'(1) = e^(-1)
y'(1) = 1/e

The equation of a line is y -y1 = m(x -x1), where m is the slope and (x1,y1) is a point on the line.
y - 1/e = (1/e)(x-1)
y - 1/e = (1/e)x - 1/e
Add 1/e to each side
y = (1/e)x or y = x/e

Answer 2

Are you asking for the tangent line to y = x^2 * e^(-x) at Point (1, 1/e)? Otherwise I do not understand what your question is. Maybe other people have the same trouble.
First, observe that point (1, 1/e) is on the curve y = x^2 * e^(-x). Therefore, you need to find the derivative of y. When the point (1, 1/e) is not on the given curve, the solution strategy would be completely different, since derivative may not be the simplest way to approach. But here it is:
y'(x) = 2x * e^(-x) - x^2 * e^(-x)
y'(1) = 2 * e^(-1) - e^(-1) = 1/e
Hence the tangent line is: y - 1/e = (1/e)(x - 1)
or simply: y = x/e