ここで勉強すれば数学検定１級の壁は超えられるか。

[math]f\left( x,y\right) =\left( 1-y^{2}\right) \tan ^{-1}\left( x+y\right)[/math]　とおくと

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

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

[math]\begin{aligned}\dfrac {\partial ^{3}f}{\partial x^{2}\partial y}\left( x,y\right) =\left( -2y\right) \dfrac {-2\left( x+y\right) }{\left( \left( x+y\right) ^{2}+1\right) ^{2}} +\left( 1-y^{2}\right) \dfrac {-2}{\left( \left( x+y\right) ^{2}+1\right) ^{2}}+\left( 1-y^{2}\right) \dfrac {8\left( x+y\right) ^{2}}{\left( \left( x+y\right) ^{2}+1\right) ^{3}}\end{aligned}[/math]

この式に

[math]x=\dfrac {1}{2},y=\dfrac {1}{2}[/math]を代入すると

[math]\dfrac {\partial ^{3}f}{\partial x^{2}\partial y}\left( \dfrac {1}{2},\dfrac {1}{2}\right) =\left( -1\right) \cdot \dfrac {-2}{\left( 1^{2}+1\right) ^{2}}[/math]

[math]+\left( 1-\dfrac {1}{4}\right) \dfrac {-2}{\left( 1^{2}+1\right) ^{2}}+\left( 1-\dfrac {1}{4}\right) \dfrac {8}{\left( 1^{2}+1\right) ^{3}}[/math]

[math]=\dfrac {1}{2}-\dfrac {3}{8}+\dfrac {3}{4}=\dfrac {7}{8}[/math]・・・答え