ここで勉強すれば数学検定１級の壁は超えられるか。

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

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

$\dfrac {\partial w}{\partial x}=\dfrac {2x}{\left( x^{2}+y^{2}+z^{2}\right) }$

$\dfrac {\partial ^{2}w}{\partial x^{2}}=\dfrac {2\left( x^{2}+y^{2}+z^{2}\right) -2x\cdot 2x}{\left( x^{2}+y^{2}+z^{2}\right) ^{2}}=\dfrac {2\left( -x^{2}+y^{2}+z^{2}\right) }{\left( x^{2}+y^{2}+z^{2}\right) ^{2}}$

同様に　$\dfrac {\partial ^{2}w}{\partial y^{2}},\dfrac {\partial ^{2}w}{\partial z^{2}}$　を計算すると

$\dfrac {\partial ^{2}w}{\partial y^{2}}=\dfrac {2\left( x^{2}-y^{2}+z^{2}\right) }{\left( x^{2}+y^{2}+z^{2}\right) ^{2}}$　

$\dfrac {\partial ^{2}w}{\partial z^{2}}=\dfrac {2\left( x^{2}+y^{2}-z^{2}\right) }{\left( x^{2}+y^{2}+z^{2}\right) ^{2}}$

$\dfrac {\partial ^{2}w}{\partial x^{2}}+\dfrac {\partial ^{2}w}{\partial y^{2}}+\dfrac {\partial ^{2}w}{\partial z^{2}}=\dfrac {2\left( x^{2}+y^{2}+z^{2}\right) }{\left( x^{2}+y^{2}+z^{2}\right) ^{2}}$

$\dfrac {2}{x^{2}+y^{2}+z^{2}}$・・・答え