# Science:Math Exam Resources/Courses/MATH104/December 2010/Question 01 (h)/Solution 1

When ${\displaystyle x=1}$, ${\displaystyle x\ln x=1\cdot 0=0}$ so we know the point ${\displaystyle (1,0)}$ is on the line.
If ${\displaystyle y=x\ln x}$ then ${\displaystyle {\frac {dy}{dx}}=(x\ln x)'=\ln x+x\cdot {\frac {1}{x}}=\ln x+1}$ and ${\displaystyle {\frac {dy}{dx}}{\bigg |}_{x=1}=1}$ so the slope of the line is ${\displaystyle 1}$.
The equation of the tangent line is then ${\displaystyle y-0=1(x-1)}$, by the point-slope formula, or ${\displaystyle \color {blue}y=x-1}$.