ここで勉強すれば数学検定１級の壁は超えられるか。

全微分可能な２変数関数ｆ（ｘ，ｙ）い対して、ｇ（rcosθ，rsinθ)とおくとき

[math]\dfrac {\partial f}{\partial x}=\dfrac {\partial g}{\partial r}\cdot a\left( r,\theta \right) +\dfrac {\partial g}{\partial \theta }\cdot b\left( r,\theta \right)[/math] 　

[math]\dfrac {\partial f}{\partial y}=\dfrac {\partial f}{\partial r}\cdot c\left( r,\theta \right) +\dfrac {\partial g}{\partial \theta }\cdot d\left( r,\theta \right)[/math]

を満たすときの関数のとき

[math]a\left( r,\theta \right) ,b\left( r,\theta \right) ,c\left( r,\theta \right) ,d\left( r,\theta \right)[/math]　を求める。

[math]r^{2}=x^{2}+y^{2},\tan \theta =\dfrac {y}{x}[/math]

[math]2r\dfrac {\partial r}{\partial x}=2x,\dfrac {1}{\cos ^{2}}\dfrac {\partial \theta }{\partial x}=-\dfrac {y}{x^{2}}[/math]

[math]\dfrac {\partial r}{\partial x}=\cos \theta ,\dfrac {\partial \theta }{\partial x}=-\dfrac {\sin \theta }{r}[/math]

[math]\dfrac {\partial f}{\partial x}=\dfrac {\partial g}{\partial r}\dfrac {\partial r}{\partial x}+\dfrac {\partial g}{\partial \theta }\dfrac {\partial \theta }{\partial x}[/math]

最初の式と比較すると

[math]a\left( r,\theta \right) =\cos \theta ,b\left( r,\theta \right) =-\dfrac {\sin \theta }{r}[/math]・・・答え

同様に

[math]2r\dfrac {\partial r}{\partial x}=2x,\dfrac {\partial \theta }{\partial y}=\dfrac {1}{x}[/math]

[math]c\left( r,\theta \right) =\sin \theta ,d\left( r,\theta \right) =\dfrac {\cos \theta }{r}[/math]・・・答え