Question

determine the derivative y' at the point (1,0)
y=ln(x^2+y^2)

Answer 1

[ln(inside)]'=(inside)'/inside

Answer 2

y' = (2x + 2yy')/(x^2+y^2) right?

Answer 3

yes!

Answer 4

solve for y'

Answer 5

y' = 2x/(x^2 +y^2 - 2x) .. if i did it right...which is a big if :)

Answer 6

that last 2x should be a 2y...

Answer 7

y' = 2x/(x^2 + y^2 - 2y) (x=1,y=0)

y' = 2/(1) = 2 ??