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Answer to Question #1911 in Calculus for Bujju

Question #1911
Find an equation of the tangent line to the curve at the point (1, 1).
y = ln (xe^x^2)
Expert's answer
The equation of the tangent line is
y-yo = f'(xo) (x-xo)
f'(y) = (ex^2 + 2x*xex^2)/(xex^2) = (1+ 2x2)/x
f(1) = 3
y -1 = 3(x-1)
y = 3x -2.

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