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

Passo 1

O enunciado pede as derivadas parciais de
f(x,y) = \bigl(x^{2} + y^{2}\bigr)\,\ln\bigl(x^{2}+y\bigr).
Vamos usar a regra do produto e a regra da cadeia. Definimos u(x,y)=x^{2}+y^{2} e v(x,y)=\ln(x^{2}+y).

Passo 2

Para \displaystyle \frac{\partial f}{\partial x} aplicamos a regra do produto:
\frac{\partial f}{\partial x} = \frac{\partial u}{\partial x}\,v + u\,\frac{\partial v}{\partial x}.
Calculemos cada termo:
\frac{\partial u}{\partial x} = 2x,
\frac{\partial v}{\partial x} = \frac{1}{x^{2}+y}\,\frac{\partial (x^{2}+y)}{\partial x} = \frac{2x}{x^{2}+y}.
Logo,
\frac{\partial f}{\partial x} = 2x\,\ln\bigl(x^{2}+y\bigr) + \bigl(x^{2}+y^{2}\bigr)\,\frac{2x}{x^{2}+y}.

Passo 3

Para \displaystyle \frac{\partial f}{\partial y} repetimos o processo:
\frac{\partial f}{\partial y} = \frac{\partial u}{\partial y}\,v + u\,\frac{\partial v}{\partial y}.
Calculando,
\frac{\partial u}{\partial y} = 2y,
\frac{\partial v}{\partial y} = \frac{1}{x^{2}+y}\,\frac{\partial (x^{2}+y)}{\partial y} = \frac{1}{x^{2}+y}.
Portanto,
\frac{\partial f}{\partial y} = 2y\,\ln\bigl(x^{2}+y\bigr) + \frac{x^{2}+y^{2}}{x^{2}+y}.

Resposta

\displaystyle \frac{\partial f}{\partial x} =2x\,\ln\bigl(x^{2}+y\bigr) +\frac{2x\,(x^{2}+y^{2})}{x^{2}+y}. \displaystyle \frac{\partial f}{\partial y} =2y\,\ln\bigl(x^{2}+y\bigr) +\frac{x^{2}+y^{2}}{x^{2}+y}.

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