Calcule as derivadas parciais das seguintes funções:
f(x, y)=\sin \left(\sqrt{2 x^{3}+y^{2}}\right)

Passo 1

O enunciado pede as derivadas parciais de
f(x,y)=\sin\bigl(\sqrt{2x^{3}+y^{2}}\bigr).
O conteúdo aqui é derivada parcial e regra da cadeia para funções compostas: derivada de \sin(u) e derivada de u=(2x^3+y^2)^{1/2}.

Passo 2

Para \displaystyle \frac{\partial f}{\partial x}, defina
u = \sqrt{2x^3+y^2}=(2x^3+y^2)^{1/2}.
Então
\frac{\partial f}{\partial x} = \cos(u)\,\frac{\partial u}{\partial x}.
Calculando
\frac{\partial u}{\partial x} = \tfrac12(2x^3+y^2)^{-1/2}\cdot 6x^2 = \frac{3x^2}{\sqrt{2x^3+y^2}}.
Logo
\frac{\partial f}{\partial x} = \cos\!\bigl(\sqrt{2x^3+y^2}\bigr)\,\frac{3x^2}{\sqrt{2x^3+y^2}}.

Passo 3

Para \displaystyle \frac{\partial f}{\partial y}, de modo análogo:
\frac{\partial f}{\partial y} = \cos(u)\,\frac{\partial u}{\partial y}, \quad \frac{\partial u}{\partial y} = \tfrac12(2x^3+y^2)^{-1/2}\cdot 2y = \frac{y}{\sqrt{2x^3+y^2}}.
Portanto
\frac{\partial f}{\partial y} = \cos\!\bigl(\sqrt{2x^3+y^2}\bigr)\,\frac{y}{\sqrt{2x^3+y^2}}.

Resposta

\frac{\partial f}{\partial x} = \frac{3x^2}{\sqrt{2x^3+y^2}}\, \cos\!\bigl(\sqrt{2x^3+y^2}\bigr),
\frac{\partial f}{\partial y} = \frac{y}{\sqrt{2x^3+y^2}}\, \cos\!\bigl(\sqrt{2x^3+y^2}\bigr).

Exercícios relacionados