Nos problemas 5-8, determine o gradiente da função dada no ponto indicado.
f(x, y)=\sqrt[2]{x^{3} y-y^{4}}\; (3; 2)

Passo 1

O enunciado pede o gradiente de
f(x,y)=\sqrt{\,x^{3}y - y^{4}\,}
no ponto (3,2).
Para resolver isso, vamos usar que, se f(x,y)=(g(x,y))^{1/2}, então as parciais vêm da regra da cadeia:
\frac{\partial f}{\partial x} =\frac12\,g^{-1/2}\,\frac{\partial g}{\partial x}, \quad \frac{\partial f}{\partial y} =\frac12\,g^{-1/2}\,\frac{\partial g}{\partial y}.

Passo 2

Definimos
g(x,y)=x^{3}y - y^{4}.
Então
\frac{\partial g}{\partial x} =3x^{2}y,
e pela regra da cadeia
\frac{\partial f}{\partial x} =\frac12\,(g(x,y))^{-1/2}\,(3x^{2}y) =\frac{3x^{2}y}{2\sqrt{\,g(x,y)\,}}.

Passo 3

De modo análogo,
\frac{\partial g}{\partial y} =x^{3} - 4y^{3},
logo
\frac{\partial f}{\partial y} =\frac12\,(g(x,y))^{-1/2}\,(x^{3}-4y^{3}) =\frac{x^{3}-4y^{3}}{2\sqrt{\,g(x,y)\,}}.

Passo 4

No ponto (3,2) temos
g(3,2)=3^{3}\cdot2 - 2^{4}=27\cdot2 - 16=54 - 16=38,
portanto \sqrt{g(3,2)}=\sqrt{38}. Assim,
\left.\frac{\partial f}{\partial x}\right|_{(3,2)} =\frac{3\cdot3^{2}\cdot2}{2\sqrt{38}} =\frac{54}{2\sqrt{38}} =\frac{27}{\sqrt{38}},
\left.\frac{\partial f}{\partial y}\right|_{(3,2)} =\frac{3^{3}-4\cdot2^{3}}{2\sqrt{38}} =\frac{27 - 32}{2\sqrt{38}} =\frac{-5}{2\sqrt{38}}.

Resposta

\nabla f(3,2) =\biggl(\frac{27}{\sqrt{38}},\;-\frac{5}{2\sqrt{38}}\biggr).

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