Question

Derivar implícitamente x^3-3axy+y^3=0​

Answer 1

SOLUCIÓN

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• Comenzaremos derivando respecto a "x"

                                 $x^3-3axy+y^3=0\\\\\\\dfrac{dy}{dx}(x^3-3axy+y^3)=\dfrac{dy}{dx}(0)\\\\\\3x^2-(3ay+3axy')+3y^2y'=0\\\\\\Despejamos\:y'\\\\\\3x^2-3ay-3axy'+3y^2y'=0\\\\\\3x^2-3ay-y'(3ax-3y^2)=0\\\\\\y'(3ax-3y^2)=3x^2-3ay\\\\\\y'=\dfrac{3x^2-3ay}{3ax-3y^2}\\\\\\\boxed{\boxed{\boldsymbol{y'=\dfrac{x^2-ay}{ax-y^2}}}}$

• Derivamos respecto a "y"

                                  $x^3-3axy+y^3=0\\\\\\\dfrac{dx}{dy}(x^3-3axy+y^3)=\dfrac{dx}{dy}(0)\\\\\\3x^2x'-(3ax+3ayx')+3y^2=0\\\\\\Despejamos\:"x"\\\\\\3x^2x'-3ax-3ax'y+3y^2=0\\\\\\x'(3x^2-3ay)-3ax+3y^2=0\\\\\\x'=\dfrac{3ax-3y^2}{3x^2-3ay}\\\\\\\boxed{\boxed{\boldsymbol{x'=\dfrac{ax-y^2}{x^2-ay}}}}$