Utilize a Regra da Cadeia para determinar as derivadas parciais indicadas. z=x^{2}+x y^{3}, x=u v^{2}+w^{3}, y=u+v e^{2} ; \frac{\partial z}{\partial u}, \frac{\partial z}{\partial v}, \frac{\partial z}{\partial w} quando u=2, \quad v=1, \quad w=0 Passo 1O enunciado dá as funções z = x^{2} + x\,y^{3},\quad x = u\,v^{2} + w^{3},\quad y = u + v\,e^{2} e pede as derivadas parciais \frac{\partial z}{\partial u}, \frac{\partial z}{\partial v} e \frac{\partial z}{\partial w} no ponto (u,v,w)=(2,1,0). Vamos usar a regra da cadeia: \frac{\partial z}{\partial u} = z_{x}\,\frac{\partial x}{\partial u} + z_{y}\,\frac{\partial y}{\partial u}, e de modo análogo para v e w. Passo 2Calculamos as derivadas parciais de z em relação a x e y: z_{x} = \frac{\partial}{\partial x}\bigl(x^{2} + x\,y^{3}\bigr) = 2x + y^{3}, z_{y} = \frac{\partial}{\partial y}\bigl(x^{2} + x\,y^{3}\bigr) = 3\,x\,y^{2}. Passo 3Agora as derivadas de x e y em relação a u,v,w: \frac{\partial x}{\partial u} = v^{2}, \quad \frac{\partial x}{\partial v} = 2\,u\,v, \quad \frac{\partial x}{\partial w} = 3\,w^{2}, \frac{\partial y}{\partial u} = 1, \quad \frac{\partial y}{\partial v} = e^{2}, \quad \frac{\partial y}{\partial w} = 0. Passo 4No ponto (u,v,w)=(2,1,0) temos x = 2\cdot1^{2} + 0^{3} = 2, \quad y = 2 + 1\cdot e^{2} = 2 + e^{2}, z_{x}\bigl|_{(2,1,0)} = 2\cdot2 + (2+e^{2})^{3} = 4 + (2+e^{2})^{3}, z_{y}\bigl|_{(2,1,0)} = 3\cdot2\,(2+e^{2})^{2} = 6\,(2+e^{2})^{2}, \frac{\partial x}{\partial u}\Bigl|=1,\quad \frac{\partial x}{\partial v}\Bigl|=2\cdot2\cdot1=4,\quad \frac{\partial x}{\partial w}\Bigl|=0, \frac{\partial y}{\partial u}\Bigl|=1,\quad \frac{\partial y}{\partial v}\Bigl|=e^{2},\quad \frac{\partial y}{\partial w}\Bigl|=0. Passo 5Aplicando a regra da cadeia: \displaystyle\frac{\partial z}{\partial u}\Bigl|_{(2,1,0)} = z_{x}\cdot1 + z_{y}\cdot1 = \bigl[4 + (2+e^{2})^{3}\bigr] + 6\,(2+e^{2})^{2} = 4 + (2+e^{2})^{3} + 6\,(2+e^{2})^{2}, \displaystyle\frac{\partial z}{\partial v}\Bigl|_{(2,1,0)} = z_{x}\cdot4 + z_{y}\cdot e^{2} = 4\bigl[4 + (2+e^{2})^{3}\bigr] + e^{2}\cdot 6\,(2+e^{2})^{2} = 16 + 4\,(2+e^{2})^{3} + 6\,e^{2}\,(2+e^{2})^{2}, \displaystyle\frac{\partial z}{\partial w}\Bigl|_{(2,1,0)} = z_{x}\cdot0 + z_{y}\cdot0 = 0. Resposta \displaystyle \frac{\partial z}{\partial u}\Bigl|_{(2,1,0)} = 4 + (2+e^{2})^{3} + 6\,(2+e^{2})^{2}, \displaystyle \frac{\partial z}{\partial v}\Bigl|_{(2,1,0)} = 16 + 4\,(2+e^{2})^{3} + 6\,e^{2}\,(2+e^{2})^{2}, \displaystyle \frac{\partial z}{\partial w}\Bigl|_{(2,1,0)} = 0. MOSTRAR SOLUÇÃO COMPLETAExercícios relacionadosUtilize a Regra da Cadeia para determinar as derivadas parciais indicadas. u=\sqrt{r^{2}+s^{2}}, r=y+x \cos t, \quad s=x+y \operatorname{sen} t ; \frac{\partial u}{\partial x}, \frac{\partial u}{\partial y}, \frac{\partial u}{\partial t} quando x=1, \quad y=2, \quad t=0 (Logístico - não linear) Aproxime y^{\prime}(t)=y(1-y), \quad y(0)=0,1 com h=0,2 até t=1,0. Utilize a Regra da Cadeia para determinar as derivadas parciais indicadas. P=\sqrt{u^{2}+v^{2}+w^{2}}, u=x e^{y}, v=y e^{x}, w=e^{x} ; \frac{\partial P}{\partial x}, \frac{\partial P}{\partial y} \quad quando x=0, \quad y=2 Utilize as Equações 7 para determinar \partial z / \partial x e \partial z / \partial y. x^{2}+2 y^{2}+3 z^{2}=1