Đáp án:
\(d^2f(1,1)= - \dfrac12dx^2 + \dfrac12xdy^2\)
Giải thích các bước giải:
\(\begin{array}{l}
\quad f(x,y) = \arctan\dfrac xy\\
\text{Ta có:}\\
+)\quad \dfrac{\partial f}{\partial x} = \dfrac{y}{x^2 + y^2}\\
+)\quad \dfrac{\partial f}{\partial y} = - \dfrac{x}{x^2 +y^2}\\
+)\quad \dfrac{\partial^2 f}{\partial x^2}= - \dfrac{2xy}{(x^2 + y^2)^2}\\
+)\quad \dfrac{\partial^2 f}{\partial x\partial y} = \dfrac{\partial^2 f}{\partial y\partial x} = \dfrac{x^2 - y^2}{(x^2 + y^2)^2}\\
+)\quad \dfrac{\partial^2 f}{\partial y^2}= \dfrac{2xy}{(x^2 + y^2)^2}\\
+)\quad d^2f = \dfrac{\partial^2 f}{\partial x^2}dx^2 + 2\dfrac{\partial^2 f}{\partial x\partial y}dxdy + \dfrac{\partial^2 f}{\partial y^2}dy^2\\
\Rightarrow d^2f = -\dfrac{2xy}{(x^2 + y^2)^2}dx^2 + 2\cdot \dfrac{x^2 - y^2}{(x^2 + y^2)^2}dxdy + \dfrac{2xy}{(x^2 + y^2)^2}dy^2\\
\text{Tại $(1;1)$ ta được:}\\
d^2f(1,1)= - \dfrac12dx^2 + \dfrac12xdy^2
\end{array}\)
