How do you determine (d^2y)/(dx^2) given x-y=(x+y)^2?

Jun 19, 2018

$y ' ' = - \frac{8}{1 + 2 x + 2 y} ^ 3$

Explanation:

By the chain rule we get

$1 - 2 \left(x + y\right) = y ' + 2 y ' \left(x + y\right)$
so
$y ' = \frac{1 - 2 \left(x + y\right)}{1 + 2 \left(x + y\right)}$

$y ' ' = \frac{- 2 \left(1 + y '\right) \left(1 + 2 \left(x + y\right)\right) - \left(1 - 2 \left(x + y\right)\right) 2 \left(1 + y '\right)}{1 + 2 x + 2 y} ^ 2$
pluggin the term above in the second derivative we get

$y ' ' = - \frac{8}{1 + 2 x + 2 y} ^ 3$