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

**y = 2x-1**

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 + 1

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