ここで勉強すれば数学検定１級の壁は超えられるか。

[math]w=\log _{e}\left( x^{2}+y^{2}+z^{2}\right)[/math]　に対して、

[math]\dfrac {\partial ^{2}w}{\partial x^{2}}+\dfrac {\partial ^{2}w}{\partial y^{2}}+\dfrac {\partial ^{2}w}{\partial z^{2}}[/math]　の次の計算をします。

[math]\dfrac {\partial w}{\partial x}=\dfrac {2x}{\left( x^{2}+y^{2}+z^{2}\right) }[/math]

[math]\dfrac {\partial ^{2}w}{\partial x^{2}}=\dfrac {2\left( x^{2}+y^{2}+z^{2}\right) -2x\cdot 2x}{\left( x^{2}+y^{2}+z^{2}\right) ^{2}}=\dfrac {2\left( -x^{2}+y^{2}+z^{2}\right) }{\left( x^{2}+y^{2}+z^{2}\right) ^{2}}[/math]

同様に　[math]\dfrac {\partial ^{2}w}{\partial y^{2}},\dfrac {\partial ^{2}w}{\partial z^{2}}[/math]　を計算すると

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

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

[math]\dfrac {\partial ^{2}w}{\partial x^{2}}+\dfrac {\partial ^{2}w}{\partial y^{2}}+\dfrac {\partial ^{2}w}{\partial z^{2}}=\dfrac {2\left( x^{2}+y^{2}+z^{2}\right) }{\left( x^{2}+y^{2}+z^{2}\right) ^{2}}[/math]

[math]\dfrac {2}{x^{2}+y^{2}+z^{2}}[/math]・・・答え