Đáp án:
$1)\quad \begin{cases}f_x' = - \dfrac{y}{x^2 + y^2}\\f_y' = \dfrac{x}{x^2 + y^2}\end{cases}$
$2)\quad \begin{cases}f_x' = \dfrac{y}{x^2 + y^2}\\f_y' = -\dfrac{x}{x^2 + y^2}\end{cases}$
Giải thích các bước giải:
\(\begin{array}{l}
1)\quad f(x,y)= \arctan\dfrac yx\\
+)\quad f_x' = \dfrac{\left(\dfrac yx\right)'}{1 + \left(\dfrac yx\right)^2}\\
\Leftrightarrow f_x' = \dfrac{- \dfrac{y}{x^2}}{1 + \dfrac{y^2}{x^2}}\\
\Leftrightarrow f_x' = - \dfrac{y}{x^2 + y^2}\\
+)\quad f_y' = \dfrac{\left(\dfrac yx\right)'}{1 + \left(\dfrac yx\right)^2}\\
\Leftrightarrow f_y' = \dfrac{\dfrac1x}{1 + \dfrac{y^2}{x^2}}\\
\Leftrightarrow f_y' = \dfrac{x}{x^2 + y^2}\\
2)\quad f(x,y)= \arctan\dfrac xy\\
+)\quad f_x' = \dfrac{\left(\dfrac xy\right)'}{1 + \left(\dfrac xy\right)^2}\\
\Leftrightarrow f_x' = \dfrac{\dfrac{1}{y}}{1 + \dfrac{x^2}{y^2}}\\
\Leftrightarrow f_x' = \dfrac{y}{x^2 + y^2}\\
+)\quad f_y' = \dfrac{\left(\dfrac xy\right)'}{1 + \left(\dfrac xy\right)^2}\\
\Leftrightarrow f_y' = \dfrac{-\dfrac{x}{y^2}}{1 + \dfrac{x^2}{y^2}}\\
\Leftrightarrow f_y' = -\dfrac{x}{x^2 + y^2}\\
\end{array}\)
