Calcule as derivadas parciais de f
f(x, y)=\sin \left(\cos \left(\ln \left(x^{2}+y\right)\right)\right)

Passo 1

O enunciado pede para calcular as derivadas parciais de
f(x, y) = \sin\bigl(\cos\bigl(\ln(x^2 + y)\bigr)\bigr).
O conteúdo principal aqui é a aplicação da regra da cadeia em funções de duas variáveis.

Passo 2

Para organizar a regra da cadeia, definimos funções intermediárias:
u = x^2 + y,\quad v = \ln(u),\quad w = \cos(v),\quad z = \sin(w).
Assim temos f(x,y)=z em que z depende de w, que depende de v, que depende de u, que por fim depende de x e y.

Passo 3

Calculamos as derivadas de u em relação a x e a y:
\frac{\partial u}{\partial x} = 2x, \qquad \frac{\partial u}{\partial y} = 1.

Passo 4

Aplicando a derivada de v=\ln(u):
\frac{\partial v}{\partial x} = \frac{1}{u}\,\frac{\partial u}{\partial x} = \frac{2x}{x^2+y}, \qquad \frac{\partial v}{\partial y} = \frac{1}{u}\,\frac{\partial u}{\partial y} = \frac{1}{x^2+y}.

Passo 5

Para w=\cos(v), usamos \frac{d}{dv}\cos(v)=-\sin(v):
\frac{\partial w}{\partial x} = -\sin(v)\,\frac{\partial v}{\partial x} = -\sin\bigl(\ln(x^2+y)\bigr)\,\frac{2x}{x^2+y},
\frac{\partial w}{\partial y} = -\sin(v)\,\frac{\partial v}{\partial y} = -\sin\bigl(\ln(x^2+y)\bigr)\,\frac{1}{x^2+y}.

Passo 6

Finalmente, para z=\sin(w) usamos \frac{d}{dw}\sin(w)=\cos(w):
\frac{\partial f}{\partial x} = \frac{\partial z}{\partial w}\,\frac{\partial w}{\partial x} = \cos\bigl(\cos(\ln(x^2+y))\bigr)\,\biggl[-\sin\bigl(\ln(x^2+y)\bigr)\,\frac{2x}{x^2+y}\biggr],
\frac{\partial f}{\partial y} = \frac{\partial z}{\partial w}\,\frac{\partial w}{\partial y} = \cos\bigl(\cos(\ln(x^2+y))\bigr)\,\biggl[-\sin\bigl(\ln(x^2+y)\bigr)\,\frac{1}{x^2+y}\biggr].

Resposta

\frac{\partial f}{\partial x} = -\,\cos\bigl(\cos(\ln(x^2+y))\bigr)\,\sin\bigl(\ln(x^2+y)\bigr)\,\frac{2x}{x^2+y},
\frac{\partial f}{\partial y} = -\,\cos\bigl(\cos(\ln(x^2+y))\bigr)\,\sin\bigl(\ln(x^2+y)\bigr)\,\frac{1}{x^2+y}.

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