Find an equation of the tangent line to the curve y = x2 ln x at (1, 0). Write your anser in slope-intercept form.

Answer 1

The curve is y=(x^2)(ln x)

The given point is (1,0)

To find the tangent curve, you must find slope at the given point.

First, take derivative of the given curve.

dy/dx = (x^2)(1/x) + (2x)(ln x) ignoring d/dx

simplify

dy/dx = x + 2x(ln x)

plug in given x-coordinate

dy/dx = 1 + 2(1) (ln 1)

dy/dx = 1 = m = slope

plug in given point and slope into slope intercept form

y - y1 = m(x-x2)

y - 0 = 1(x-1)

your answer is

y = x - 1

Cheers!