✅ Após resolver todos os cálculos, concluímos que a derivada implícita mais geral da curva do diabo definida por y²(y² - 4) = x²(x² - 5) é:
[tex]\Large\displaystyle\text{$\begin{gathered}\boxed{\boxed{\:\:\:\bf y' = \frac{2x^{3} - 5x}{2y^{3} - 4y}\:\:\:}}\end{gathered}$}[/tex]
Seja curva do diabo:
[tex]\Large\displaystyle\text{$\begin{gathered} y^{2}(y^{2} - 4) = x^{2}(x^{2} - 5)\end{gathered}$}[/tex]
Calculando a derivada implícita da curva em termos da incógnita "x":
[tex]\Large \text {$\begin{aligned}(y^{2}(y^{2} - 4))' & = (x^{2}(x^{2} - 5))' \\(y^{4} - 4y^{2})' & = (x^{4} - 5x^{2})'\\4y^{3}\,y' - 8y\,y' & = 4x^{3} - 10x\\(4y^{3} - 8y)y' & = 4x^{3} - 10x\\y' & = \frac{4x^{3} - 10x}{4y^{3} - 8y}\\ y' & = \frac{{\!\diagup\!\!\!\!2}\cdot(2x^{3} - 5x)}{{\!\diagup\!\!\!\!2}\cdot(2y^{3}- 4y)}\\ y' & = \frac{2x^{3} - 5x}{2y^{3} - 4y}\end{aligned} $}[/tex]
✅ Portanto, a derivada implícita da curva em termos de "x" é:
[tex]\Large\displaystyle\text{$\begin{gathered} y' = \frac{2x^{3} - 5x}{2y^{3} - 4y}\end{gathered}$}[/tex]
[tex]\LARGE\displaystyle\text{$\begin{gathered} \underline{\boxed{\boldsymbol{\:\:\:Bons \:estudos!!\:\:\:Boa\: sorte!!\:\:\:}}}\end{gathered}$}[/tex]
Saiba mais:
[tex]\Large\displaystyle\text{$\begin{gathered} \underline{\boxed{\boldsymbol{\:\:\:Observe \:o\:Gr\acute{a}fico!!\:\:\:}}}\end{gathered}$}[/tex]
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✅ Após resolver todos os cálculos, concluímos que a derivada implícita mais geral da curva do diabo definida por y²(y² - 4) = x²(x² - 5) é:
[tex]\Large\displaystyle\text{$\begin{gathered}\boxed{\boxed{\:\:\:\bf y' = \frac{2x^{3} - 5x}{2y^{3} - 4y}\:\:\:}}\end{gathered}$}[/tex]
Seja curva do diabo:
[tex]\Large\displaystyle\text{$\begin{gathered} y^{2}(y^{2} - 4) = x^{2}(x^{2} - 5)\end{gathered}$}[/tex]
Calculando a derivada implícita da curva em termos da incógnita "x":
[tex]\Large \text {$\begin{aligned}(y^{2}(y^{2} - 4))' & = (x^{2}(x^{2} - 5))' \\(y^{4} - 4y^{2})' & = (x^{4} - 5x^{2})'\\4y^{3}\,y' - 8y\,y' & = 4x^{3} - 10x\\(4y^{3} - 8y)y' & = 4x^{3} - 10x\\y' & = \frac{4x^{3} - 10x}{4y^{3} - 8y}\\ y' & = \frac{{\!\diagup\!\!\!\!2}\cdot(2x^{3} - 5x)}{{\!\diagup\!\!\!\!2}\cdot(2y^{3}- 4y)}\\ y' & = \frac{2x^{3} - 5x}{2y^{3} - 4y}\end{aligned} $}[/tex]
✅ Portanto, a derivada implícita da curva em termos de "x" é:
[tex]\Large\displaystyle\text{$\begin{gathered} y' = \frac{2x^{3} - 5x}{2y^{3} - 4y}\end{gathered}$}[/tex]
[tex]\LARGE\displaystyle\text{$\begin{gathered} \underline{\boxed{\boldsymbol{\:\:\:Bons \:estudos!!\:\:\:Boa\: sorte!!\:\:\:}}}\end{gathered}$}[/tex]
Saiba mais:
[tex]\Large\displaystyle\text{$\begin{gathered} \underline{\boxed{\boldsymbol{\:\:\:Observe \:o\:Gr\acute{a}fico!!\:\:\:}}}\end{gathered}$}[/tex]