Respuesta :

vivindalka
[tex]x^2+3xy+y^3=10\\2x+3xy'+3y+2yy'=0\\(2x+3y)+(3xy'+2yy')=0\\2x+3y+(3x+2y)y'=0\\(3x+2y)y'=-2x-3y\ /\ :\ (3x+2y)\\y'=\frac{-2x-3y}{3x+2y}[/tex]


IamONLYgoodFORmath

Answer:

https://tex.z-dn.net/?f=x%5E2%2B3xy%2By%5E3%3D10%5C%5C2x%2B3xy%27%2B3y%2B2yy%27%3D0%5C%5C%282x%2B3y%29%2B%283xy%27%2B2yy%27%29%3D0%5C%5C2x%2B3y%2B%283x%2B2y%29y%27%3D0%5C%5C%283x%2B2y%29y%27%3D-2x-3y%5C%20%2F%5C%20%3A%5C%20%283x%2B2y%29%5C%5Cy%27%3D%5Cfrac%7B-2x-3y%7D%7B3x%2B2y%7D

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