How do you determine $\frac{d^{2} y}{{d x}^{2}}$ $(d^2y)/(dx^2)$ given $x - y = {(x + y)}^{2}$ $x-y=(x+y)^2$ ?

Question

1808 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-19T18:29:25

$y''=-8/(1+2x+2y)^3$

By the chain rule we get

$1-2(x+y)=y'+2y'(x+y)$
so
$y'=(1-2(x+y))/(1+2(x+y))$

$y''=(-2(1+y')(1+2(x+y))-(1-2(x+y))2(1+y'))/(1+2x+2y)^2$
pluggin the term above in the second derivative we get

$y''=-8/(1+2x+2y)^3$

How do you determine (d^2y)/(dx^2)d2ydx2(d^2y)/(dx^2) given x-y=(x+y)^2x−y=(x+y)2x-y=(x+y)^2?