Asked by Anna
Find an equation of the tangent line to the graph of y = e^(βx^2) at the point (5, 1/(e^25)).
My work:
y' = -2xe^(-x^2)
y' when x = 5 = -10e^25
y - (1/e^25) = -10e^25 * (x - 5)
y = (-10e^25) (x - 5) + (1/e^25)
This is incorrect. How?
My work:
y' = -2xe^(-x^2)
y' when x = 5 = -10e^25
y - (1/e^25) = -10e^25 * (x - 5)
y = (-10e^25) (x - 5) + (1/e^25)
This is incorrect. How?
Answers
Answered by
Steve
y' when x = 5 = -10e^-25
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