Função diferenciável - Cálculo 2

Nesse exercício vamos estudar derivadas parciais.

Sabemos que:

$$uz + y\ln(u) – u^2 + 24 = 3y^3e^{4-z^2y^2}$$

Derivando implicitamente em relação a $y$, temos:

$$\dfrac{\partial u}{\partial y}z + \ln(u) + \dfrac{y}{u}\dfrac{\partial u}{\partial y} – 2u \dfrac{\partial u}{\partial y} = 9y^2e^{4-z^2y^2}+3y^3e^{4-z^2y^2}(-2z^2y)$$

$$\left(z-2u+\dfrac{y}{u}\right)\dfrac{\partial u}{\partial y}+ \ln(u) = 3y^2 e^{4-z^2y^2} (3-2y^2z^2)$$

$$\dfrac{\partial u}{\partial y}=\dfrac{3y^2 e^{4-z^2y^2} (3-2y^2z^2)- \ln(u) }{ z-2u+\dfrac{y}{u} }$$

Para o ponto estudado:

$$\dfrac{\partial u}{\partial y}(2,1)=\dfrac{3\cdot2^2 e^{4-1^2\cdot2^2} (3-2\cdot2^2\cdot1^2)- \ln u(2,1) }{ 1-2u(2,1)+ \dfrac{2}{u(2,1)} }=\dfrac{12e^{4-4} (3-8)- \ln 1 }{ 1-2+\dfrac{2}{1} }$$

$$\dfrac{\partial u}{\partial y}(2,1)=-60$$

Derivando implicitamente em relação a $z$, temos:

$$\dfrac{\partial u}{\partial z}z+ u + \dfrac{y}{u} \dfrac{\partial u}{\partial z}-2u\dfrac{\partial u}{\partial z} = 3y^3e^{4-z^2y^2}(-2zy^2)$$

$$\left(z- 2u +\dfrac{y}{u}\right) \dfrac{\partial u}{\partial z} = -6y^5ze^{4-z^2y^2}-u$$

$$\dfrac{\partial u}{\partial z} = -\dfrac{6y^5ze^{4-z^2y^2}+u}{ z- 2u +\dfrac{y}{u}}$$

Para o ponto estudado:

$$\dfrac{\partial u}{\partial z}(2,1) = -\dfrac{6\cdot2^5\cdot1e^{4-1^2\cdot2^2}+u(2,1)}{ 1- 2u(2,1) +\dfrac{2}{u(2,1)}} = -\dfrac{192e^{4-4}+1}{ 1- 2 +\dfrac{2}{1}}$$

$$\dfrac{\partial u}{\partial z}(2,1)=-193$$

Finalmente:

$$\boxed{\dfrac{\partial u}{\partial y}(2,1)- \dfrac{\partial u}{\partial z}(2,1)=133}$$

Nesse exercício vamos estudar derivadas parciais.

Sabemos que: