3.2 正則関数

1.

微分公式と微分法を確認しよう.

(a)

$\displaystyle (z^3 - 2z^2 + 3z)'$ $\displaystyle =$ $\displaystyle (z^3)' -2(z^2)' + 3z' \ (和の導関数は導関数の和)$  
  $\displaystyle =$ $\displaystyle 3z^2 - 4z + 3  ((z^{\alpha})' = \alpha z^{\alpha - 1})$  

(b)

$\displaystyle ((z^2 +i)^{3})'$ $\displaystyle =$ $\displaystyle 3(z^2 + i)^{2}(z^2 + i)' \ (合成関数の微分法より)$  
  $\displaystyle =$ $\displaystyle 3(z^2 + i)^{2}(2z) = 6z(z^2 + i)^{2}$  

(c)

$\displaystyle (\frac{z - i}{z + i})'$ $\displaystyle =$ $\displaystyle \frac{(z-i)'(z+i) - (z-i)(z+i)'}{(z+ i)^{2}} \ (商の微分法より)$  
  $\displaystyle =$ $\displaystyle \frac{z+i - (z-i)}{(z+ i)^{2}} = \frac{2i}{(z+ i)^{2}}$  

2.

(a)

$\displaystyle (\tan{z})'$ $\displaystyle =$ $\displaystyle \frac{\sin{z}}{\cos{z}} = \frac{(\sin{z})'(\cos{z}) - (\sin{z})(\cos{z})'}{(\cos{z})^{2}} \ (商の微分法より)$  
  $\displaystyle =$ $\displaystyle \frac{\cos^{2}{z} + \sin^{2}{z}}{\cos^{2}{z}} = \frac{1}{\cos^{2}} = \sec^{2}{z}$  

(b)

$\displaystyle (\frac{1}{\cos{z}})'$ $\displaystyle =$ $\displaystyle \frac{- (\cos{z})'}{(\cos{z})^{2}} \ (商の微分法より)$  
  $\displaystyle =$ $\displaystyle \frac{\sin{z}}{\cos^{2}{z}} = \sec{z}\tan{z}$  

(c)


$\displaystyle (\sqrt{z^2 + 1})'$ $\displaystyle =$ $\displaystyle ((z^2 + 1)^{1/2})' = \frac{1}{2}(z^2 + 1)^{-1/2}(z^2 + 1)' \ (合成関数の微分法より)$  
  $\displaystyle =$ $\displaystyle z(z^2 + 1)^{-1/2}$  

(d)

$\displaystyle (\sin^{2}{z})'$ $\displaystyle =$ $\displaystyle ((\sin{z})^{2})' = 2(\sin{z})(\sin{z})' \ (合成関数の微分法より)$  
  $\displaystyle =$ $\displaystyle 2\sin{z}\cos{z}$  

(e)

$\displaystyle (\log(z^2+4i))'$ $\displaystyle =$ $\displaystyle \frac{1}{z^2 + 4i}(z^2 + 4i)' \ (合成関数の微分法より)$  
  $\displaystyle =$ $\displaystyle \frac{2z}{z^2 + 4i}$  

(f)

$\displaystyle (i^{\cos{z}})'$ $\displaystyle =$ $\displaystyle (e^{\cos{z}\log{i}})' \ (a^{z} = e^{z\log{a}}より)$  
  $\displaystyle =$ $\displaystyle e^{\cos{z}\log{i}}(\cos{z}\log{i})' \ (合成関数の微分法より)$  
  $\displaystyle =$ $\displaystyle e^{\cos{z}\log{i}}(-\sin{z}\log{i}) = -i^{\cos{z}}\sin{z}\log{i}$  

(g)

$\displaystyle (\sin^{-1}(z-i))'$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{1 - (z-i)^2}}(z-i)' \ (合成関数の微分法より)$  
  $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{1 - (z-i)^2}}$  

(h)

$\displaystyle (\log(z + \sqrt{z^2 + 1}))'$ $\displaystyle =$ $\displaystyle \frac{1}{z + \sqrt{z^2 + 1}}(z+\sqrt{z^2 + 1})' \ (合成関数の微分法より)$  
  $\displaystyle =$ $\displaystyle \frac{1}{z + \sqrt{z^2 + 1}}(1 + \frac{z}{\sqrt{z^2 + 1}})$  
  $\displaystyle =$ $\displaystyle \frac{1}{z + \sqrt{z^2 + 1}}(\frac{\sqrt{z^2 + 1} + z}{\sqrt{z^2 + 1}}$  
  $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{z^2 + 1}}$  

(i)

$\displaystyle (\log(\sin^{-1}{z}))'$ $\displaystyle =$ $\displaystyle \frac{1}{\sin^{-1}{z}}(\sin^{-1}{z})' \ (合成関数の微分法より)$  
  $\displaystyle =$ $\displaystyle \frac{1}{\sin^{-1}{z}}(\frac{1}{\sqrt{1 - z^2}})$  
  $\displaystyle =$ $\displaystyle \frac{1}{\sin^{-1}{z}\sqrt{1 - z^2}}$  

(j)

$\displaystyle (z^{z})'$ $\displaystyle =$ $\displaystyle (e^{z\log{z}})' = e^{z\log{z}}(z\log{z})'$  
  $\displaystyle =$ $\displaystyle e^{z\log{z}}(\log{z} + z\frac{1}{z}) = z^{z}(\log{z} + 1)$