Question

find the derivative of x^2 y^2+x^3+y=8 at (2,0)

Answer 1

2xy^2+x^2(2yy')+3x^2+y'=0 can you solve this for y'

Answer 2

y'(2yx^2+1)=-3x^2-2xy^2

Answer 3

y'=[-3x^2-2xy^2]/[2yx^2+1]

Answer 4

thanks so now do I just plug in 2 for x and 0 for y?

Answer 5

yes

Answer 6

The derivative \[2xy^2 + 2x^2y(y') + 3x^2 + y' = 0\] So rearrange giving: \[y'(1+ 2x^2y) = -2xy^2 - 3x^2\] as myinaya said

Answer 7

thanks guys!