Asked by Scooby10
Find the slope of the tangent line to the curve (x + (e^-x))^2 at (0,1)
Answers
Answered by
Steve
the slope of the tangent line is the derivative of y=(x+e^-x)^2 at x=0
y' = 2(x + e^-x)(1 - e^-x)
y'(0) = 2(0+1)(1-1) = 0
y' = 2(x + e^-x)(1 - e^-x)
y'(0) = 2(0+1)(1-1) = 0
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