Question

Derivação implicitamente, obtenha dy/dx

X^3+y^3=4xy+1, (4,1)

Answer 1

$\displaystyle i)~~~~x^3+y^3=4xy+1\\\\ii)~~~\frac{d}{dx}\left(x^3+y^3\right)=\frac{d}{dx}\left(4xy+1\right)\\\\iii)~~3x^2+3y^2\frac{dy}{dx}=4\left(y+x\frac{dy}{dx}\right)\\\\iv)~~3x^2-4y=\frac{dy}{dx}\left(4x-3y^2\right)\\\\v)~~\boxed{\frac{dy}{dx}=\frac{3x^2-4y}{4x-3y^2}}$

calcular no ponto:
$\displaystyle \frac{dy}{dx}(x,y)=\frac{3x^2-4y}{x-3y^2}\\\\\frac{dy}{dx}(4,1)=\frac{3(4)^2-4}{4(4)-3}=\frac{3\cdot 16-4}{13}\approx\boxed{3,38}$

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Answer 2

Obtenha $\frac{dy}{dx}$
$x^{3} + y^{3} =4xy+1$
$3 x^{2} \frac{dx}{dx} +3 y^{2} \frac{dy}{dx} =4(1.\frac{dx}{dx}.y+x.1.[tex]3 x^{2}+3 y^{2} \frac{dy}{dx}=4y+4x\frac{dy}{dx}$)+0[/tex]
$\frac{dy}{dx}(3 y^{2} -4x)=4y-3 x^{2}$
$\frac{dy}{dx}=\frac{4y-3 x^{2}}{3 y^{2} -4x}$