3) Determine as derivadas parciais f_(x),f_(y),f_(xx),f_(xy),f_(yx)ef_(yy) de a) f(x,t)=sqrt (x)lnt b) f(x,y)=x^4y^3+8x^2y c) f(x,y)=(x)/((x+y)^2) d) f(x,y)=tan(xy)

Vamos calcular as derivadas parciais solicitadas para cada função.

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### **a) $f(x,t) = \sqrt{x} \ln t$**

1. $f_x = \frac{\partial}{\partial x} (\sqrt{x} \ln t) = \frac{1}{2\sqrt{x}} \ln t$
2. $f_t = \frac{\partial}{\partial t} (\sqrt{x} \ln t) = \sqrt{x} \cdot \frac{1}{t} = \frac{\sqrt{x}}{t}$

Agora, as segundas derivadas:
3. $f_{xx} = \frac{\partial}{\partial x} \left( \frac{1}{2\sqrt{x}} \ln t \right) = -\frac{1}{4x^{3/2}} \ln t$
4. $f_{xt} = f_{tx} = \frac{\partial}{\partial t} \left( \frac{1}{2\sqrt{x}} \ln t \right) = \frac{1}{2\sqrt{x}} \cdot \frac{1}{t} = \frac{1}{2t\sqrt{x}}$
5. $f_{tt} = \frac{\partial}{\partial t} \left( \frac{\sqrt{x}}{t} \right) = \sqrt{x} \cdot \left( -\frac{1}{t^2} \right) = -\frac{\sqrt{x}}{t^2}$

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### **b) $f(x,y) = x^4 y^3 + 8x^2 y$**

1. $f_x = \frac{\partial}{\partial x} (x^4 y^3 + 8x^2 y) = 4x^3 y^3 + 16x y$
2. $f_y = \frac{\partial}{\partial y} (x^4 y^3 + 8x^2 y) = 3x^4 y^2 + 8x^2$

Segundas derivadas:
3. $f_{xx} = \frac{\partial}{\partial x} (4x^3 y^3 + 16x y) = 12x^2 y^3 + 16y$
4. $f_{xy} = f_{yx} = \frac{\partial}{\partial y} (4x^3 y^3 + 16x y) = 12x^3 y^2 + 16x$
5. $f_{yy} = \frac{\partial}{\partial y} (3x^4 y^2 + 8x^2) = 6x^4 y$

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### **c) $f(x,y) = \frac{x}{(x+y)^2}$**

1. $f_x = \frac{\partial}{\partial x} \left( \frac{x}{(x+y)^2} \right) = \frac{(x+y)^2 - 2x(x+y)}{(x+y)^4} = \frac{(x+y) - 2x}{(x+y)^3} = \frac{-x + y}{(x+y)^3}$
2. $f_y = \frac{\partial}{\partial y} \left( \frac{x}{(x+y)^2} \right) = \frac{-2x(x+y)}{(x+y)^4} = \frac{-2x}{(x+y)^3}$

Segundas derivadas:
3. $f_{xx} = \frac{\partial}{\partial x} \left( \frac{-x + y}{(x+y)^3} \right) = \frac{-1(x+y)^3 - 3(-x+y)(1)}{(x+y)^6} = \frac{-1(x+y)^3 + 3(x-y)}{(x+y)^4}$
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