Calculus: Early Transcendentals, Chapter 3, 3.6, Section 3.6, Problem 34

The equation of the tangent line to the curve y = x^2*ln x , at the point (1,0) is the following, such that:
f(x) - f(x_0) = f'(x_0)(x - x_0)
You need to put f(x) = y, f(x_0) = 0, x_0 = 1
You need to evaluate f'(1), hence, you need to find the derivative of the function, using the product rule, such that:
f'(x) = (x^2)'*ln x + x^2*(ln x)'
f'(x) = 2x*ln x + x^2*(1/x)
f'(x) = 2x*ln x + x
Replacing 1 for x, yields:
f'(1) = 2*ln 1 + 1
f'(1) = 2*0 + 1 => f'(1) = 1
Replacing the values in the equation of tangent line, yields:
y - 0 = 1*(x - 1)
y = x - 1
Hence, evaluating the equation of the tangent line to the given curve, at the point (1,0), yields y = x - 1.