Find the derivative of the function at P_{0} in the direction of \mathbf{A}.
h(x, y)=\tan ^{-1}(y / x)+\sqrt{3} \sin ^{-1}(x y / 2), \quad P_{0}(1,1)\(,\mathbf{A}=3 \mathbf{i}-2 \mathbf{j})

Passo 1

O enunciado pede a derivada direcional de h(x,y) no ponto P_{0}(1,1) na direção do vetor \mathbf{A} = 3\,\mathbf{i} - 2\,\mathbf{j}.
Conteúdo necessário: derivada direcional de função de duas variáveis, que se obtém como produto escalar entre o gradiente de h e o vetor unitário na direção de \mathbf{A}.

Passo 2

Vamos achar o gradiente \nabla h(x,y) = \bigl(\partial h/\partial x,\;\partial h/\partial y\bigr).
Para a parte \tan^{-1}(y/x):
\frac{\partial}{\partial x}\tan^{-1}\!\Bigl(\frac{y}{x}\Bigr) = \frac{1}{1+(y/x)^{2}}\cdot\Bigl(-\frac{y}{x^{2}}\Bigr) = -\frac{y}{x^{2}+y^{2}},
\frac{\partial}{\partial y}\tan^{-1}\!\Bigl(\frac{y}{x}\Bigr) = \frac{1}{1+(y/x)^{2}}\cdot\frac{1}{x} = \frac{x}{x^{2}+y^{2}}.
Para a parte \sqrt{3}\,\sin^{-1}(xy/2), defina v=xy/2:
\frac{\partial}{\partial x}\bigl[\sqrt{3}\,\sin^{-1}(v)\bigr] = \sqrt{3}\,\frac{1}{\sqrt{1-v^{2}}}\cdot\frac{\partial v}{\partial x} = \sqrt{3}\,\frac{1}{\sqrt{1-(xy/2)^{2}}}\cdot\frac{y}{2},
\frac{\partial}{\partial y}\bigl[\sqrt{3}\,\sin^{-1}(v)\bigr] = \sqrt{3}\,\frac{1}{\sqrt{1-(xy/2)^{2}}}\cdot\frac{x}{2}.
Logo,
\nabla h(x,y) = \Bigl(-\frac{y}{x^{2}+y^{2}} + \sqrt{3}\,\frac{y/2}{\sqrt{1-(xy/2)^{2}}},\; \frac{x}{x^{2}+y^{2}} + \sqrt{3}\,\frac{x/2}{\sqrt{1-(xy/2)^{2}}}\Bigr).

Passo 3

Avaliaremos \nabla h em (x,y)=(1,1). Aqui x^{2}+y^{2}=2 e xy/2=1/2, de modo que
\sqrt{1-(xy/2)^{2}} = \sqrt{1-\tfrac{1}{4}} = \frac{\sqrt{3}}{2}, \quad \frac{1}{\sqrt{1-(xy/2)^{2}}} = \frac{2}{\sqrt{3}}.
Assim,
\frac{\partial h}{\partial x}(1,1) = -\frac{1}{2} + \sqrt{3}\,\frac{1/2}{\sqrt{3}/2} = -\frac{1}{2} + \sqrt{3}\,\frac{1}{2}\cdot\frac{2}{\sqrt{3}} = -\frac{1}{2} + 1 = \frac{1}{2},
\frac{\partial h}{\partial y}(1,1) = \frac{1}{2} + \sqrt{3}\,\frac{1/2}{\sqrt{3}/2} = \frac{1}{2} + 1 = \frac{3}{2}.
Portanto
\nabla h(1,1) = \Bigl(\tfrac{1}{2},\,\tfrac{3}{2}\Bigr).

Passo 4

O vetor unitário na direção de \mathbf{A}=(3,-2) é
\mathbf{u} = \frac{1}{\|\mathbf{A}\|}(3,-2) = \frac{1}{\sqrt{3^{2}+(-2)^{2}}}(3,-2) = \frac{1}{\sqrt{13}}\,(3,-2).

Passo 5

A derivada direcional de h em P_{0} na direção de \mathbf{A} é
D_{\mathbf{u}}h(1,1) = \nabla h(1,1)\cdot \mathbf{u} = \Bigl(\tfrac{1}{2},\,\tfrac{3}{2}\Bigr)\cdot \Bigl(\tfrac{3}{\sqrt{13}},\, -\tfrac{2}{\sqrt{13}}\Bigr) = \frac{1}{2}\cdot\frac{3}{\sqrt{13}} + \frac{3}{2}\cdot\Bigl(-\frac{2}{\sqrt{13}}\Bigr) = -\frac{3}{2\sqrt{13}}.

Resposta

D_{\mathbf{A}}h(1,1) \;=\; -\frac{3}{2\sqrt{13}}.

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