Respuesta :

Given: 3y cos x = x² + y²

Define [tex]y' = \frac{dy}{dx} [/tex]

Then by implicit differentiation, obtain
3y' cos(x) - 3y sin(x) = 2x + 2y y'
y' [3 cos(x) - 2y] = 2x + 3y sinx)

Answer:
[tex] \frac{dy}{dx} = \frac{2x+3y \, sin(x)}{3cos(x)-2y} [/tex]
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