Calcule a derivada parcial de primeira ordem:
f(x, y)=x \cos (2 x+3 y)^{2}

Passo 1

O enunciado pede as derivadas parciais de primeira ordem de
f(x,y)=x\cos\bigl((2x+3y)^2\bigr).
O que vamos usar: regra do produto (para x\cdot\cos(\cdot)) e regra da cadeia (para \cos((2x+3y)^2)).

Passo 2

Calcular \displaystyle\frac{\partial f}{\partial x}. Aplicando regra do produto:
\frac{\partial f}{\partial x} =\frac{\partial}{\partial x}(x)\,\cos\bigl((2x+3y)^2\bigr) + x\,\frac{\partial}{\partial x}\Bigl[\cos\bigl((2x+3y)^2\bigr)\Bigr].
Como \displaystyle\frac{\partial}{\partial x}(x)=1 e usando cadeia em
\frac{\partial}{\partial x}\Bigl[\cos\bigl((2x+3y)^2\bigr)\Bigr] =-\sin\bigl((2x+3y)^2\bigr)\,\frac{\partial}{\partial x}\bigl[(2x+3y)^2\bigr],
com
\frac{\partial}{\partial x}\bigl[(2x+3y)^2\bigr] =2(2x+3y)\cdot 2=4(2x+3y),
segue
\frac{\partial f}{\partial x} =\cos\bigl((2x+3y)^2\bigr) -4x(2x+3y)\sin\bigl((2x+3y)^2\bigr).

Passo 3

Calcular \displaystyle\frac{\partial f}{\partial y}. Agora x é tratado como constante, então
\frac{\partial f}{\partial y} =x\,\frac{\partial}{\partial y}\Bigl[\cos\bigl((2x+3y)^2\bigr)\Bigr] =-x\sin\bigl((2x+3y)^2\bigr)\,\frac{\partial}{\partial y}\bigl[(2x+3y)^2\bigr].
E
\frac{\partial}{\partial y}\bigl[(2x+3y)^2\bigr] =2(2x+3y)\cdot 3=6(2x+3y),
logo
\frac{\partial f}{\partial y} =-6x(2x+3y)\sin\bigl((2x+3y)^2\bigr).

Resposta

\displaystyle\frac{\partial f}{\partial x} =\cos\bigl((2x+3y)^2\bigr) -4x(2x+3y)\sin\bigl((2x+3y)^2\bigr)
\displaystyle\frac{\partial f}{\partial y} =-6x(2x+3y)\sin\bigl((2x+3y)^2\bigr)

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