Question

How do I find the second derivative implicitly of 2y-x+xy=6. I figured out the first derivative is (1-y)/(2+x) but I don't know what to do next. Please show what steps to use

Answer 1

y" = 2(y - 1)/(2 + x)²

Explanation:

2y-x+xy=6

2y' - 1 + (xy' + y) = 0

Differentiate again wrt x

2y" + (xy" + y') + y' = 0

2y" + xy" + 2y' = 0

y"(2 + x) = -2y'

y"(2 + x) = -2(1 - y)/(2 + x)

y" = -2(1 - y)/(2 + x)²

y" = 2(y - 1)/(2 + x)²