Considere f(x, y)=x^{4} y^{5}-2 x^{3} y+3.
(a) Obtenha o vetor gradiente \vec{\nabla} f(x, y);
(b) Calcule a derivada direcional de f, D_{\vec{u}} f(P), sendo P=(1,-1) e \vec{u}=\frac{\sqrt{2}}{2} \vec{i}+\frac{\sqrt{2}}{2} \vec{j}.

Passo 1

O enunciado pede (a) o vetor gradiente de f(x,y)=x^{4}y^{5}-2x^{3}y+3 e (b) a derivada direcional de f no ponto P=(1,-1) na direção do vetor unitário \displaystyle \vec{u}=\frac{\sqrt{2}}{2}\,\vec{i}+\frac{\sqrt{2}}{2}\,\vec{j}. Vamos usar cálculo de derivadas parciais para obter o gradiente e depois aplicar
D_{\vec{u}}f(P)=\vec{\nabla}f(P)\cdot\vec{u}.

Passo 2

Calcular as derivadas parciais de f.
\frac{\partial f}{\partial x} = \frac{\partial}{\partial x}\bigl(x^{4}y^{5}-2x^{3}y+3\bigr) =4x^{3}y^{5}-6x^{2}y,
\frac{\partial f}{\partial y} = \frac{\partial}{\partial y}\bigl(x^{4}y^{5}-2x^{3}y+3\bigr) =5x^{4}y^{4}-2x^{3}.
Portanto
\vec{\nabla}f(x,y)=\bigl(4x^{3}y^{5}-6x^{2}y,\;5x^{4}y^{4}-2x^{3}\bigr).

Passo 3

Avaliar o gradiente em P=(1,-1):
\frac{\partial f}{\partial x}(1,-1) =4\cdot1^{3}\cdot(-1)^{5}-6\cdot1^{2}\cdot(-1) =-4+6=2,
\frac{\partial f}{\partial y}(1,-1) =5\cdot1^{4}\cdot(-1)^{4}-2\cdot1^{3} =5-2=3.
Logo
\vec{\nabla}f(1,-1)=(2,3).

Passo 4

Usar a definição de derivada direcional:
D_{\vec{u}}f(1,-1) =\vec{\nabla}f(1,-1)\cdot\vec{u} =(2,3)\cdot\Bigl(\tfrac{\sqrt2}{2},\tfrac{\sqrt2}{2}\Bigr) =2\cdot\tfrac{\sqrt2}{2}+3\cdot\tfrac{\sqrt2}{2} =\tfrac{5\sqrt2}{2}.

Resposta

(a) \displaystyle \vec{\nabla}f(x,y)=\bigl(4x^{3}y^{5}-6x^{2}y,\;5x^{4}y^{4}-2x^{3}\bigr).
(b) \displaystyle D_{\vec{u}}f(1,-1)=\frac{5\sqrt2}{2}.

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