Calcule as derivadas parciais \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial^{2} f}{\partial x^{2}}, \frac{\partial^{2} f}{\partial y^{2}}, \frac{\partial^{2} f}{\partial x \partial y} para as funções abaixo:
f(x y)=\operatorname{sen}\left(x^{2}+y\right)

Passo 1

O enunciado fornece a função
f(x,y)=\sin\bigl(x^{2}+y\bigr).
O que se pede são as derivadas parciais de primeira ordem \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, de segunda ordem \frac{\partial^{2}f}{\partial x^{2}}, \frac{\partial^{2}f}{\partial y^{2}} e a derivada mista \frac{\partial^{2}f}{\partial x\partial y}. Vamos usar a regra da cadeia e, quando necessário, a regra do produto.

Passo 2

Para \displaystyle \frac{\partial f}{\partial x}: defina a função interna
u(x,y)=x^{2}+y.
Então f(x,y)=\sin(u). Pela regra da cadeia,
\frac{\partial f}{\partial x} =\cos(u)\,\frac{\partial u}{\partial x} =\cos\bigl(x^{2}+y\bigr)\cdot 2x =2x\cos\bigl(x^{2}+y\bigr).

Passo 3

Para \displaystyle \frac{\partial f}{\partial y}: ainda com u=x^{2}+y,
\frac{\partial f}{\partial y} =\cos(u)\,\frac{\partial u}{\partial y} =\cos\bigl(x^{2}+y\bigr)\cdot 1 =\cos\bigl(x^{2}+y\bigr).

Passo 4

Para \displaystyle \frac{\partial^{2}f}{\partial x^{2}}: partindo de \displaystyle \frac{\partial f}{\partial x}=2x\cos(u), aplique a regra do produto e em seguida a cadeia:
\frac{\partial^{2}f}{\partial x^{2}} =\frac{\partial}{\partial x}\bigl(2x\cos(u)\bigr) =2\cos(u)+2x\bigl(-\sin(u)\bigr)\frac{\partial u}{\partial x} =2\cos\bigl(x^{2}+y\bigr)-4x^{2}\sin\bigl(x^{2}+y\bigr).

Passo 5

Para \displaystyle \frac{\partial^{2}f}{\partial y^{2}}: partindo de \displaystyle \frac{\partial f}{\partial y}=\cos(u),
\frac{\partial^{2}f}{\partial y^{2}} =\frac{\partial}{\partial y}\bigl(\cos(u)\bigr) =-\sin(u)\,\frac{\partial u}{\partial y} =-\sin\bigl(x^{2}+y\bigr).

Passo 6

Para \displaystyle \frac{\partial^{2}f}{\partial x\partial y}: derive \displaystyle \frac{\partial f}{\partial x}=2x\cos(u) em relação a y:
\frac{\partial^{2}f}{\partial x\partial y} =2x\bigl(-\sin(u)\bigr)\frac{\partial u}{\partial y} =-2x\sin\bigl(x^{2}+y\bigr).

Resposta

\displaystyle \frac{\partial f}{\partial x}=2x\cos\bigl(x^{2}+y\bigr) \displaystyle \frac{\partial f}{\partial y}=\cos\bigl(x^{2}+y\bigr) \displaystyle \frac{\partial^{2}f}{\partial x^{2}} =2\cos\bigl(x^{2}+y\bigr)-4x^{2}\sin\bigl(x^{2}+y\bigr) \displaystyle \frac{\partial^{2}f}{\partial y^{2}} =-\sin\bigl(x^{2}+y\bigr) \displaystyle \frac{\partial^{2}f}{\partial x\partial y} =-2x\sin\bigl(x^{2}+y\bigr)

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