Question

calcule a derivada implícita x^3+y^3=-6xy​

Answer 1

Resposta:

3x²+3y² *dy/dx=-6y -6xdy/dx

3y² *dy/dx +6xdy/dx =-6y-3x²

dy/dx* (3y²+6x)= 6y-3x²

dy/dx* (y²+2x)= 2y-x²

dy/dx =(2y-x²)/(y²+2x)

Answer 2

✅ Tendo finalizado os cálculos, concluímos que a derivada implícita da curva definida implicitamente em termos da incógnita "x" é:

                   $\Large\displaystyle\text{ \begin{gathered}\boxed{\boxed{\:\:\:\bf y' = \frac{-2y - x^{2}}{y^{2} + 2x}\:\:\:}}\end{gathered} }$

 

Seja a função definida implicitamente:

                          $\Large\displaystyle\begin{gathered} x^{3} + y^{3} = -6xy\end{gathered}$

Calculando a derivada implícita, temos da incógnita "x":

   $\Large \text { \begin{aligned}(x^{3} + y^{3})' & = (-6xy)'\\3\cdot x^{3 - 1} + 3\cdot y^{2 - 1}y' & = -6\cdot(1\cdot x^{1 - 1}\cdot y + x\cdot 1\cdot y^{1 - 1}y')\\3x^{2} + 3y^{2}y' & = -6\cdot(y + xy')\\3x^{2} + 3y^{2}y'& = -6y - 6xy'\\ 3y^{2}y' + 6xy' & = -6y - 3x^{2}\\(3y^{2} + 6x)y' & = -6y - 3x^{2}\\y' & = \frac{-6y - 3x^{2}}{3y^{2} + 6x}\\y' & = \frac{3\cdot(-2y - x^{2})}{3\cdot(y^{2} + 2x)}\\y' & = \frac{-2y - x^{2}}{y^{2} + 2x}\end{aligned} }$

✅ Portanto, a derivada implícita em termos da incógnita "x" é:

                                       $\Large\displaystyle\text{ \begin{gathered} y' = \frac{-2y - x^{2}}{y^{2} + 2x}\end{gathered} }$

$\LARGE\displaystyle\text{ \begin{gathered} \underline{\boxed{\boldsymbol{\:\:\:Bons \:estudos!!\:\:\:Boa\: sorte!!\:\:\:}}}\end{gathered} }$

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$\Large\displaystyle\text{ \begin{gathered} \underline{\boxed{\boldsymbol{\:\:\:Observe \:o\:Gr\acute{a}fico!!\:\:\:}}}\end{gathered} }$