Question

1) isole o y da equação

a) 2y^2+xy=x^2+3

b)x^2+xy-y^2=3

depois de isolar o y, tem que derivar o y, usando y' em função de de X

RESOLUÇÃO já derivado !
A) Y' = 1/4(-1+- ( 9X / RAIZ(9X^2+24)))

B)Y' = 1/2(1+-( 5x / raiz( 5x^2 - 12 )

Me ajudem por favor, essas duas questões são as únicas que não estou conseguindo isolar literalmente o y!!!

Answer 1

$\boxed{\boxed{\dfrac{d}{dx}(cf(x))=c\dfrac{d}{dx}f(x)}}\\\\\\\boxed{\boxed{\dfrac{d}{dx}\sqrt{x}=\dfrac{d}{dx}x^{1/2}=\dfrac{1}{2\sqrt{x}}}}\\\\\\\boxed{\boxed{\dfrac{d}{dx}f(g(x))=\dfrac{dy}{du}\dfrac{du}{dx}=f'(g(x))\cdot g'(x)}}$
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Tem como isolar y sim! Veja:

a)

$2y^{2}+xy=x^{2}+3\\\\2y^{2}+xy-(x^{2}+3)=0$

Podemos isolar y resolvendo essa equação do segundo grau (com variável y) e encontrar y em função de x

$2y^{2}+xy-(x^{2}+3)=0\\\\\Delta=b^{2}-4ac\\\Delta=x^{2}-4\cdot2\cdot[-(x^{2}+3)]\\\Delta=x^{2}+8(x^{2}+3)\\\Delta=x^{2}+8x^{2}+24\\\Delta=9x^{2}+24\\\\y=\dfrac{-b\pm\sqrt{\Delta}}{2a}=\dfrac{-x\pm\sqrt{9x^{2}+24}}{2\cdot2}=\dfrac{-x\pm\sqrt{9x^{2}+24}}{4}\\\\\\\boxed{\boxed{y=\dfrac{1}{4}(-x\pm\sqrt{9x^{2}+24})}}$

Derivando y em relação a x:

$\dfrac{dy}{dx}=\dfrac{d}{dx}\dfrac{1}{4}(-x\pm\sqrt{9x^{2}+24})\\\\\\\dfrac{dy}{dx}=\dfrac{1}{4}\left(\dfrac{d}{dx}(-x)\pm\dfrac{d}{dx}\sqrt{9x^{2}+24}\right)\\\\\\\dfrac{dy}{dx}=\dfrac{1}{4}\left(-1\pm\dfrac{1}{2\sqrt{9x^{2}+24}}\cdot\dfrac{d}{dx}(9x^{2}+24)}\right)\\\\\\\dfrac{dy}{dx}=\dfrac{1}{4}\left(-1\pm\dfrac{1}{2\sqrt{9x^{2}+24}}\cdot18x\right)\\\\\\\boxed{\boxed{\dfrac{dy}{dx}=\dfrac{1}{4}\left(-1\pm\dfrac{9x}{\sqrt{9x^{2}+24}}\right)}}$
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b)

$x^{2}+xy-y^{2}=3\\\\-y^{2}+xy+x^{2}-3=0\\\\-y^{2}+xy+(x^{2}-3)=0$

Resolvendo a equação para acharmos y em função de x:

$-y^{2}+xy+(x^{2}-3)=0\\\\\Delta=b^{2}-4ac\\\Delta=x^{2}-4\cdot(-1)\cdot(x^{2}-3)\\\Delta=x^{2}+4(x^{2}-3)\\\Delta=x^{2}+4x^{2}-12\\\Delta=5x^{2}-12\\\\y=\dfrac{-b\pm\sqrt{\Delta}}{2a}=\dfrac{-x\pm\sqrt{5x^{2}-12}}{2\cdot(-1)}=\dfrac{-x\pm\sqrt{5x^{2}-12}}{(-2)}\\\\\\\boxed{\boxed{y=-\dfrac{1}{2}\left(-x\pm\sqrt{5x^{2}-12}\right)}}$

Derivando y em função de x:

$y=-\dfrac{1}{2}\left(-x\pm\sqrt{5x^{2}-12}\right)\\\\\\\dfrac{dy}{dx}=\dfrac{d}{dx}\left[-\dfrac{1}{2}(-x\pm\sqrt{5x^{2}-12})\right]\\\\\\\dfrac{dy}{dx}=-\dfrac{1}{2}\left(\dfrac{d}{dx}(-x)\pm\dfrac{d}{dx}\sqrt{5x^{2}-12}\right)\\\\\\\dfrac{dy}{dx}=-\dfrac{1}{2}\left(-1\pm\dfrac{1}{2\sqrt{5x^{2}-12}}\cdot\dfrac{d}{dx}(5x^{2}-12)}\right)\\\\\\\dfrac{dy}{dx}=-\dfrac{1}{2}\left(-1\pm\dfrac{1}{2\sqrt{5x^{2}-12}\cdot10x}\right)$

$\boxed{\boxed{\dfrac{dy}{dx}=-\dfrac{1}{2}\left(-1\pm\dfrac{5x}{\sqrt{5x^{2}-12}}\right)}}$