cálculo diferencial

Use regra da cadeia e observe se está diferenciando para x ou y.

\[\vec u=\nabla f(1,1)=\left(\left.\dfrac{\partial f}{\partial x}\right|_{(1,1)},\left.\dfrac{\partial f}{\partial y}\right|_{(1,1)}\right)\]

Vamos então determinar as derivadas parciais:

\[\dfrac{\partial f}{\partial x}=\arctan\dfrac{x}y+\dfrac{x}y\cdot\dfrac{1}{1+\left(\dfrac{x}y\right)^2}=\arctan\dfrac{x}y+\dfrac{xy}{x^2+y^2}\]

\[\dfrac{\partial f}{\partial y}=x\cdot\dfrac{1}{1+\left(\dfrac{x}y\right)^2}\cdot\left(-\dfrac{x}{y^2}\right)=-\dfrac{x^2}{x^2+y^2}\]

Substituindo o ponto desejado, temos:

\[\left.\dfrac{\partial f}{\partial x}\right|_{(1,1)}=\arctan\dfrac11+\dfrac{1\cdot1}{1^2+1^2}=\dfrac\pi4+\dfrac12\]

\[\left.\dfrac{\partial f}{\partial y}\right|_{(1,1)}=-\dfrac{1^2}{1^2+1^2}=\dfrac12\]

Temos então a direção desejada:

\[\vec u=\nabla f(1,1)=\left(\dfrac\pi4+\dfrac12,\dfrac12\right)\]

Para a derivada, temos:

\[\left.\dfrac{df}{d\vec u}\right|_{(1,1)}=\nabla f(1,1)\cdot\dfrac{\vec u}{|\vec u|}\]

Mas \(\nabla f(1,1)=\vec u\), então:

\[\left.\dfrac{df}{d\vec u}\right|_{(1,1)}=\vec u\cdot\dfrac{\vec u}{|\vec u|}=\dfrac{\vec u^2}{|\vec u|}\$=\dfrac{|\vec u|^2}{|\vec u|}=|\vec u|\]

Usando a expressão do módulo de um vetor, com o vetor já obtido, temos:

\[\left.\dfrac{df}{d\vec u}\right|_{(1,1)}=\sqrt{\left(\dfrac\pi4+\dfrac12\right)^2+\left(\dfrac12\right)^2}=\sqrt{\dfrac{\pi^2}{16}+\dfrac\pi4+\dfrac14+\dfrac14}\]

Finalmente:

\[\boxed{\left.\dfrac{df}{d\vec u}\right|_{(1,1)}=\dfrac14\sqrt{\pi^2+4\pi+8}}\]