Question

find an equation of the tangent line to the graph of y=xlnx at the point (1,0)

Answer 1

\[y'=(1) \ln(x)+x \frac{1}{x}=\ln(x)+1\]

Answer 2

\[y'|_{x=1}=\ln(1)+1=0+1=1\]

Answer 3

\[y=x+b\]

Answer 4

we need to find b the y-intercept

Answer 5

we know a point on this line: (1,0)

Answer 6

0=1+b=>
-1=b=>
b=-1

Answer 7

\[y=x-1\]