Question

Derivada implícita de: 2x+3y/x^2+y^2=9

Answer 1

Olá

Temos que aplicar a regra do quociente

$\displaystyle \mathsf{ (\frac{f}{g})' = \frac{f'\cdot g -f\cdot g'}{g^2} }$


$\displaystyle \mathsf{ \frac{2x+3y}{x^2+y^2}=9 }$

Derivando implicitamente


$\displaystyle \mathsf{ \frac{(2+3y')\cdot(x^2+y^2)~-~(2x+3y)\cdot(2x+2y\cdot y')}{(x^2+y^2)^2}=0 }\\\\\\\text{Aplica a distributiva}\\\\\\\mathsf{ \frac{(2x^2+2y^2+3x^2y'+3y^2y')~-~(4x^2+4xyy'+6xy+6y^2y')}{(x^2+y^2)^2}=0 }\\\\\\\text{Passa o denominador para o outro lado, multiplicando}\\\\\\\mathsf{(2x^2+2y^2+3x^2y'+3y^2y')-(4x^2+4xyy'+6xy+6y^2y')=0\cdot (x^2+y^2)^2 }\\\\\\\mathsf{(2x^2+2y^2+3x^2y'+3y^2y')-(4x^2+4xyy'+6xy+6y^2y')=0}\\\\\\\mathsf{2x^2+2y^2+3x^2y'+3y^2y'-4x^2-4xyy'-6xy-6y^2y'=0}$

Passe todos os termos que não possui y' para o outro lado

$\displaystyle\mathsf{3x^2y'+3y^2y'-4xyy'-6y^2y'=-2x^2-2y^2+4x^2+6xy}\\\\\\\mathsf{3x^2y'+\diagup\!\!\!\!3y^2y'-4xyy'-\diagup\!\!\!\!6y^2y'=-\diagup\!\!\!\!2x^2-2y^2+\diagup\!\!\!\!\!4x^2+6xy}\\\\\\\mathsf{3x^2y'-3y^2y'-4xyy'=-2y^2+2x^2+6xy}\\\\\\\text{Poe o y' em evidencia}\\\\\\\mathsf{y'(3x^2-3y^2-4xy)=-2y^2+2x^2+6xy}\\\\\\\text{Isola o y'}\\\\\\\boxed{\mathsf{y'= \frac{2x^2-2y^2+6xy}{3x^2-3y^2-4xy} }}$