implicit differentiation of x^3 y + x y^3 = 30

[gopher]

ok
i need to find T.L and N.L for
x^3y+xy^3=30

my first step i got

(3x^2*y)+(x^3*y') + (y^3)+(x*3y^2*y') = 0

is that right ? do i need to use product rule like that?

 

[galactus]

Yep.

\(\displaystyle \L\\x^{3}y+xy^{3}=30\)

\(\displaystyle \L\\x^{3}\frac{dy}{dx}+3x^{2}y+3xy^{2}\frac{dy}{dx}+y^{3}=0\)

\(\displaystyle \L\\\frac{dy}{dx}=\frac{-y^{3}-3x^{2}y}{x^{3}+3xy^{2}}\)