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Answer to Question #1690 in Calculus for Bujju

Question #1690
Find an equation of the tangent line to the curve at the point (1, 1).
y = ln (xex2)
Expert's answer
Let's find the first derivation of the function:
f' = (xex^2)'/(xex^2) = (1+x2)/x
The equation of the tangent line can be written as follows:
y = f(x0) + f'(x0) (x-x0)
In our case:
y = ln (1*e) + (1 + 12)/1 * (x-1) = 1 + 2(x-1) = 2x -1
y = 2x-1

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