ここで勉強すれば数学検定１級の壁は超えられるか。

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

$\dfrac {\partial f}{\partial x}\left( x,y\right) =\left( 1-y^{2}\right) \dfrac {1}{\left( x+y\right) ^{2}+1}$

$\dfrac {\partial ^{2}f}{\partial x^{2}}\left( x,y\right) =\left( 1-y^{2}\right) \dfrac {-2\left( x+y\right) }{\left( \left( x+y\right) ^{2}+1\right) ^{2}}$

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

この式に

$x=\dfrac {1}{2},y=\dfrac {1}{2}$を代入すると

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

$+\left( 1-\dfrac {1}{4}\right) \dfrac {-2}{\left( 1^{2}+1\right) ^{2}}+\left( 1-\dfrac {1}{4}\right) \dfrac {8}{\left( 1^{2}+1\right) ^{3}}$

$=\dfrac {1}{2}-\dfrac {3}{8}+\dfrac {3}{4}=\dfrac {7}{8}$・・・答え