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) = (e^{x^2} + 2x*xe^{x^2})/(xe^{x^2}) = (1+ 2x^{2})/x

f(1) = 3

y -1 = 3(x-1)

y = 3x -2.

y-yo = f'(xo) (x-xo)

f'(y) = (e

f(1) = 3

y -1 = 3(x-1)

y = 3x -2.

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