find the equation of the tangent line to the graph of y=2xe^-x+e^2x at (0,1)

Question

find the equation of the tangent line to the graph of y=2xe^-x+e^2x at (0,1)

Answer 1

Find the first derivative of y(x) - this will be the slope of the line at the xy point (0,1).

y'(x) = -2e^-x(x-1) + 2e^2x = m (the slope of the line)

Solve for x = 0: y'(0) = 2 + 2 = 4 = m

Equation of tangent line: y = mx + b

Solve for b by using the xy point (0,1): 1 = 4(0) + b; b = 1

Solution: y = 4x + 1