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

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

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

を満たすときの関数のとき

$a\left( r,\theta \right) ,b\left( r,\theta \right) ,c\left( r,\theta \right) ,d\left( r,\theta \right)$　を求める。

$r^{2}=x^{2}+y^{2},\tan \theta =\dfrac {y}{x}$

$2r\dfrac {\partial r}{\partial x}=2x,\dfrac {1}{\cos ^{2}}\dfrac {\partial \theta }{\partial x}=-\dfrac {y}{x^{2}}$

$\dfrac {\partial r}{\partial x}=\cos \theta ,\dfrac {\partial \theta }{\partial x}=-\dfrac {\sin \theta }{r}$

$\dfrac {\partial f}{\partial x}=\dfrac {\partial g}{\partial r}\dfrac {\partial r}{\partial x}+\dfrac {\partial g}{\partial \theta }\dfrac {\partial \theta }{\partial x}$

最初の式と比較すると

$a\left( r,\theta \right) =\cos \theta ,b\left( r,\theta \right) =-\dfrac {\sin \theta }{r}$・・・答え

同様に

$2r\dfrac {\partial r}{\partial x}=2x,\dfrac {\partial \theta }{\partial y}=\dfrac {1}{x}$

$c\left( r,\theta \right) =\sin \theta ,d\left( r,\theta \right) =\dfrac {\cos \theta }{r}$・・・答え