Calcule \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}:
f(x, y)=\sqrt{x+2 y} e^{\left(-x+y^{2}\right)}

Passo 1

O enunciado pede as derivadas parciais de
f(x,y)=\sqrt{x+2y}\;e^{(-x+y^2)}
ou seja, vamos aplicar a regra do produto e a regra da cadeia, pois f é produto de
u(x,y)=(x+2y)^{1/2} \quad\text{e}\quad v(x,y)=e^{(-x+y^2)}.

Passo 2

Para \frac{\partial f}{\partial x} usamos
\frac{\partial f}{\partial x} =\frac{\partial u}{\partial x}\,v+u\,\frac{\partial v}{\partial x}.
Calculamos primeiro
\frac{\partial u}{\partial x} =\frac12\,(x+2y)^{-1/2},
e
\frac{\partial v}{\partial x} =e^{(-x+y^2)}\cdot\frac{d}{dx}(-x+y^2) =-\,e^{(-x+y^2)}.
Substituindo,
\frac{\partial f}{\partial x} =\frac12\,(x+2y)^{-1/2}\,e^{(-x+y^2)} -\,(x+2y)^{1/2}\,e^{(-x+y^2)}.

Passo 3

Fatorando e^{(-x+y^2)} e (x+2y)^{-1/2},
\frac{\partial f}{\partial x} =e^{(-x+y^2)}\,(x+2y)^{-1/2} \Bigl(\tfrac12-(x+2y)\Bigr).

Passo 4

Para \frac{\partial f}{\partial y} fazemos
\frac{\partial f}{\partial y} =\frac{\partial u}{\partial y}\,v+u\,\frac{\partial v}{\partial y}.
Temos
\frac{\partial u}{\partial y} =\frac12\,(x+2y)^{-1/2}\cdot2 =(x+2y)^{-1/2},
e
\frac{\partial v}{\partial y} =e^{(-x+y^2)}\cdot\frac{d}{dy}(-x+y^2) =2y\,e^{(-x+y^2)}.
Logo
\frac{\partial f}{\partial y} =(x+2y)^{-1/2}\,e^{(-x+y^2)} +2y\,(x+2y)^{1/2}\,e^{(-x+y^2)}.

Passo 5

Fatorando também em \frac{\partial f}{\partial y}:
\frac{\partial f}{\partial y} =e^{(-x+y^2)}\,(x+2y)^{-1/2} \bigl(1+2y\,(x+2y)\bigr).

Resposta

\frac{\partial f}{\partial x} =e^{(-x+y^2)}\,(x+2y)^{-1/2)\!\bigl(\tfrac12-(x+2y)\bigr), \quad \frac{\partial f}{\partial y} =e^{(-x+y^2)}\,(x+2y)^{-1/2)\!\bigl(1+2y\,(x+2y)\bigr).

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