Calcule \nabla \operatorname{arctg}\left(\frac{y}{x}\right) e \nabla \operatorname{arctg}\left(\frac{x}{y}\right).

Passo 1

O enunciado pede para calcular os gradientes de \displaystyle \operatorname{arctg}\!\bigl(\tfrac{y}{x}\bigr) e \displaystyle \operatorname{arctg}\!\bigl(\tfrac{x}{y}\bigr).
Vamos usar a regra da cadeia para gradiente e a derivada
\frac{d}{du}\,\operatorname{arctg}(u)=\frac{1}{1+u^2}.

Passo 2

Começando por \displaystyle f(x,y)=\operatorname{arctg}\!\bigl(\tfrac{y}{x}\bigr),
defina u(x,y)=\tfrac{y}{x}. Então
\nabla f=\frac{1}{1+u^2}\,\nabla u.
Calculando \nabla u:
\frac{\partial}{\partial x}\Bigl(\tfrac{y}{x}\Bigr)=-\frac{y}{x^2}, \quad \frac{\partial}{\partial y}\Bigl(\tfrac{y}{x}\Bigr)=\frac{1}{x},
logo
\nabla u = \Bigl(-\tfrac{y}{x^2},\,\tfrac{1}{x}\Bigr).

Passo 3

Substituindo em \nabla f e usando
1+u^2=1+\frac{y^2}{x^2}=\frac{x^2+y^2}{x^2},
temos
\nabla \operatorname{arctg}\!\Bigl(\tfrac{y}{x}\Bigr) =\frac{1}{1+u^2}\nabla u =\frac{x^2}{\,x^2+y^2\,}\Bigl(-\tfrac{y}{x^2},\,\tfrac{1}{x}\Bigr) =\Bigl(-\tfrac{y}{x^2+y^2},\,\tfrac{x}{x^2+y^2}\Bigr).

Passo 4

Agora para \displaystyle g(x,y)=\operatorname{arctg}\!\bigl(\tfrac{x}{y}\bigr),
defina v(x,y)=\tfrac{x}{y}. Então
\nabla g=\frac{1}{1+v^2}\,\nabla v.
Calculando \nabla v:
\frac{\partial}{\partial x}\Bigl(\tfrac{x}{y}\Bigr)=\frac{1}{y}, \quad \frac{\partial}{\partial y}\Bigl(\tfrac{x}{y}\Bigr)=-\frac{x}{y^2},
logo
\nabla v=\Bigl(\tfrac{1}{y},\,-\tfrac{x}{y^2}\Bigr).

Passo 5

Substituindo em \nabla g e usando
1+v^2=1+\frac{x^2}{y^2}=\frac{x^2+y^2}{y^2},
obtém-se
\nabla \operatorname{arctg}\!\Bigl(\tfrac{x}{y}\Bigr) =\frac{1}{1+v^2}\nabla v =\frac{y^2}{\,x^2+y^2\,}\Bigl(\tfrac{1}{y},\,-\tfrac{x}{y^2}\Bigr) =\Bigl(\tfrac{y}{x^2+y^2},\,-\tfrac{x}{x^2+y^2}\Bigr).

Resposta

\displaystyle \nabla \operatorname{arctg}\!\Bigl(\tfrac{y}{x}\Bigr) =\Bigl(-\tfrac{y}{x^2+y^2},\,\tfrac{x}{x^2+y^2}\Bigr) \displaystyle \nabla \operatorname{arctg}\!\Bigl(\tfrac{x}{y}\Bigr) =\Bigl(\tfrac{y}{x^2+y^2},\,-\tfrac{x}{x^2+y^2}\Bigr).

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