what is the equation of the line tangent to the curve y=x^2lnx when x=2?

Answer 1

Supposing that
y=x^(2lnx) = e^2lnx*lnx= e^[2ln^2(x)]
y´=e^2ln^2(x)*2(2lnx*1/x)
At x=2 y´= 2^2ln2 *ln2 and the tangent is
y-2^2ln2 = 2^2ln2*ln2(x-2)


If you meant y=x^2*ln x
y´= 2xlnx+x and at x=2 y´=4ln2+2
so
y-4ln2=(4ln2+2)(x-2)
y= (4ln2+2)x -4ln2-4

Answer 2

the tangent line of a curve is given by the derivative of this curve.