Question

find the slope of the tangent to the graph
3xy - 2x + 3y^2 = 5
at the point (2,1)

Answer 1

so take the derivative

\((3xy)'-2x'+(3y^2)'=0
\\3xy'+3y-2+6yy'=0\\y'=\frac{-3y+2}{3x+6y}\)

Answer 2

same

Answer 3

plug in for x,y...

Answer 4

ok hold on

Answer 5

wait I plug the x and y they give me for -3y+2 / 3x+6y

Answer 6

\(\frac{-3(1)+2}{3(2)+6(1)}=\frac{-1}{12}\)

Answer 7

ok yes , that is what I did