What is the equation of the line tangent to the graph of fx

What is the equation of the line tangent to the graph of f(x) = xlnx at the point (1,0) ?

Solution

y - f(1) = f\'(1)(x-1)

<=> y-0 = f\'(1)(x-1)

<=> y = f\'(1)(x-1) (*)

We have:

f(x) = xlnx

=> f\'(x) = x.(1/x) + lnx.1

=> f\'(1) = 1 + 0 = 1

From (*) => y = x - 1