I tried doing this myself but I'm really not sure my answer makes sense. :/
find dy/dx by implicit differentiation
square root (x+y)=1 +x^2 y^2
Hello,
Consider y as a function with respect to x.
So you have $\displaystyle \sqrt{x+y}$
What is its derivative ? It's $\displaystyle \frac{(x+y)'}{2 \sqrt{x+y}}$ (using chain rule)
What is $\displaystyle (x+y)'$ ? It's $\displaystyle (x)'+(y)'=1+y'$
So $\displaystyle (\sqrt{x+y})'=\frac{1+y'}{2 \sqrt{x+y}}$
You have $\displaystyle 1+x^2 y^2$. What I advise you is to write $\displaystyle 1+(xy)^2$
What is its derivative ? Using chain rule, we have : $\displaystyle (1)'+2(xy)'(xy)=2(xy)'(xy)$
What is $\displaystyle (xy)'$ ? Using product rule, we have : $\displaystyle y+xy'$
So $\displaystyle (1+(xy)^2)'=2(y+xy')(xy)$
$\displaystyle \boxed{\frac{1+y'}{2 \sqrt{x+y}}=2(y+xy')(xy)}$
Solve for $\displaystyle y'=\frac{dy}{dx}$
******** chain rule says : $\displaystyle f(g(x))=g'(x) f'(g(x))$