「おもな乗法的微積分による結果とその解釈」の拡張　微分編

$$\begin{align}\sqrt[dx]{d{}^x a}&=\exp(\frac{d}{dx}\ln (a^{a^{x-1}}))\\ &=\exp(\frac{d}{dx}a^{x-1}\ln(a)) \\&=\exp(a^{x-1}{(\ln(a))}^2) \\&=\exp(\ln(a^{a^{x-1}}){(\ln(a))}) \\&=\exp(\ln({}^x a){(\ln(a))}) \\&=({}^x a)^{\ln(a)} \end{align}$$

$$\begin{align}\sqrt[dx]{d{}^a x}&=\exp(\frac{d}{dx}\ln (x^{x^{a-1}}))\\ &=\exp(\frac{d}{dx}x^{a-1}\ln(x)) \\&=\exp(x^{a-2}{({(a-1)}\ln(x)+1)}) \end{align}$$

$$\begin{align}\operatorname{slog}^*_a(x)&=\exp(\frac{d}{dx}\ln (1+\log_a{(\log_a(x))}))\\ &=\exp(\frac{d(1+\log_a(\log_a(x)))/dx}{1+\log_a(\log_a(x))})\\ &=\exp(\frac{1/((\ln(a))^2x\log_a(x))}{1+\log_a(\log_a(x))})\\ &=\exp(\frac{1/(x\ln(x)\ln(a))}{1+\log_a(\log_a(x))})\\ &=\exp(\frac{1}{(x\ln(x)\ln(a))(1+\log_a(\log_a(x)))})\end{align}$$

$$\begin{align}\sin^*(x)&=\exp(\frac{d}{dx}\ln(\sin (x)))\\&=\exp(\frac{\cos(x)}{\sin(x)})\\&=e^{\cot(x)}\end{align}$$

$$\begin{align}\cos^*(x)&=\exp(\frac{d}{dx}\ln(\cos (x)))\\&=\exp(-\frac{\sin(x)}{\cos(x)})\\&=e^{-\tan(x)}\end{align}$$

$$\begin{align}\tan^*(x)&=\exp(\frac{d}{dx}\ln(\tan (x)))\\&=\exp(\frac{\sec^2(x)}{\tan(x)})\\&=\exp(\frac{1}{\sin(x)\cos(x)})\\&=\exp(\frac{2}{\sin(2x)})\end{align}$$

$$\begin{align}\cot^*(x)&=\exp(\frac{d}{dx}\ln(\cot (x)))\\&=\exp(-\frac{\csc^2(x)}{\cot(x)})\\&=\exp(\frac{1}{-\sin(x)\cos(x)})\\&=\exp(-\frac{2}{\sin(2x)})\end{align}$$

$$\begin{align}\sec^*(x)&=\exp(\frac{d}{dx}\ln(\sec (x)))\\&=\exp(\frac{\sec(x)\tan(x)}{\sec(x)})\\&=e^{\tanh(x)}\end{align}$$

$$\begin{align}\csc^*(x)&=\exp(\frac{d}{dx}\ln(\csc (x)))\\&=\exp(-\frac{\csc(x)\cot(x)}{\cot(x)})\\&=e^{-\cot(x)}\end{align}$$

$$\begin{align}\arcsin^*(x)&=\exp(\frac{d}{dx}\ln(\arcsin (x)))\\&=\exp(\frac{1/\sqrt{1-x^2}}{\arcsin(x)})\\&=\exp(\frac{1}{\sqrt{1-x^2}\arcsin(x)})\end{align}$$

$$\begin{align}\arccos^*(x)&=\exp(\frac{d}{dx}\ln(\arccos (x)))\\&=\exp(-\frac{1/\sqrt{1-x^2}}{\arccos(x)})\\&=\exp(-\frac{1}{\sqrt{1-x^2}\arccos(x)})\end{align}$$

$$\begin{align}\arctan^*(x)&=\exp(\frac{d}{dx}\ln(\arctan (x)))\\&=\exp(\frac{1/{(1+x^2)}}{\arctan(x)})\\&=\exp(\frac{1}{(1+x^2)\arctan(x)})\end{align}$$

$$\begin{align}\arccot^*(x)&=\exp(\frac{d}{dx}\ln(\arccot (x)))\\&=\exp(-\frac{1/{(1+x^2)}}{\arccot(x)})\\&=\exp(-\frac{1}{(1+x^2)\arccot(x)})\end{align}$$

$$\begin{align}\arcsec^*(x)&=\exp(\frac{d}{dx}\ln(\arcsec (x)))\\&=\exp(\frac{1/({x^2\sqrt{1-1/x^2}})}{\arcsec(x)})\\&=\exp(\frac{1}{x^2\sqrt{1-1/x^2}\arcsec(x)})\end{align}$$

$$\begin{align}\arccsc^*(x)&=\exp(-\frac{d}{dx}\ln(\arccsc (x)))\\&=\exp(-\frac{1/({x^2\sqrt{1-1/x^2}})}{\arccsc(x)})\\&=\exp(-\frac{1}{x^2\sqrt{1-1/x^2}\arccsc(x)})\end{align}$$

$$\begin{align}\sinh^*(x)&=\exp(\frac{d}{dx}\ln(\sinh (x)))\\&=\exp(\frac{\cosh(x)}{\sinh(x)})\\&=e^{\coth(x)}\end{align}$$

$$\begin{align}\cosh^*(x)&=\exp(\frac{d}{dx}\ln(\cosh (x)))\\&=\exp(\frac{\sinh(x)}{\cosh(x)})\\&=e^{\tanh(x)}\end{align}$$

$$\begin{align}\tanh^*(x)&=\exp(\frac{d}{dx}\ln(\tanh (x)))\\&=\exp(\frac{\sech^2(x)}{\tanh(x)})\\&=\exp(\frac{1}{\sinh(x)\cosh(x)})\\&=\exp(\frac{2}{\sinh(2x)})\end{align}$$

$$\begin{align}\coth^*(x)&=\exp(\frac{d}{dx}\ln(\coth (x)))\\&=\exp(-\frac{\csch^2(x)}{\coth(x)})\\&=\exp(-\frac{1}{\sinh(x)\cosh(x)})\\&=\exp(-\frac{2}{\sinh(2x)})\end{align}$$

$$\begin{align}\sech^*(x)&=\exp(\frac{d}{dx}\ln(\sech (x)))\\&=\exp(-\frac{\sech(x)\tanh(x)}{\sech(x)})\\&=e^{-\tanh(x)}\end{align}$$

$$\begin{align}\csch^*(x)&=\exp(\frac{d}{dx}\ln(\csch (x)))\\&=\exp(-\frac{\csch(x)\coth(x)}{\coth(x)})\\&=e^{-\coth(x)}\end{align}$$

$$\begin{align}\operatorname{arsinh}^*(x)&=\exp(\frac{d}{dx}\ln{(\operatorname{arsinh} (x))})\\&=\exp(\frac{1/\sqrt{x^2+1}}{\operatorname{arsinh}(x)})\\&=\exp(\frac{1}{\sqrt{x^2+1}\operatorname{arsinh}(x)})\end{align}$$

$$\begin{align}\operatorname{arcosh}^*(x)&=\exp(\frac{d}{dx}\ln{(\operatorname{arcosh} (x))})\\&=\exp(\frac{1/\sqrt{x^2-1}}{\operatorname{arcosh}(x)})\\&=\exp(\frac{1}{\sqrt{x^2-1}\operatorname{arcosh}(x)})\end{align}$$

$$\begin{align}\operatorname{artanh}^*(x)&=\exp(\frac{d}{dx}\ln{(\operatorname{artanh}(x))})\\&=\exp(\frac{1/{(1-x^2)}}{\operatorname{artanh}(x)})\\&=\exp(\frac{1}{(1-x^2)\operatorname{artanh}(x)})\end{align}$$

$$\begin{align}\operatorname{arcoth}^*(x)&=\exp({\frac{d}{dx}\ln(\operatorname{arcoth}(x))})\\&=\exp(\frac{1/{(1-x^2)}}{\operatorname{arcoth}(x)})\\&=\exp(\frac{1}{(1-x^2)\operatorname{arcoth}(x)})\end{align}$$

$$\begin{align}\operatorname{arsech}^*(x)&=\exp({\frac{d}{dx}\ln(\operatorname{arsech}(x))})\\&=\exp(-\frac{1/{x(x+1)\sqrt{(1-x)/(1+x)}}}{\operatorname{arsech}(x)})\\&=\exp(-\frac{1}{x(x+1)\sqrt{(1-x)/(1+x)}\operatorname{arsech}(x)})\end{align}$$

$$\begin{align}\operatorname{arcsch}^*(x)&=\exp({\frac{d}{dx}\ln(\operatorname{arcsch}(x))})\\&=\exp(-\frac{1/({x^2\sqrt{1+1/x^2}})}{\operatorname{arcsch}(x)})\\&=\exp(-\frac{1}{x^2\sqrt{1+1/x^2}\operatorname{arcsch}(x)})\end{align}$$

$$ \operatorname{gd}(x)= \int_{0}^{x}\sech(t)dt$$なので、
$$\begin{align}\operatorname{gd}^*(x)&=\exp({\frac{d}{dx}\ln(\operatorname{gd}(x))})\\&=\exp(\frac{\sech(x)}{\operatorname{gd}(x)})\end{align}$$

$$\operatorname{gd}^{-1}(x)= \int_{0}^{x}\sec(t)dt$$
なので、
$$\begin{align}{\operatorname{gd}^{-1}}^*(x)&=\exp({\frac{d}{dx}\ln(\operatorname{gd}^{-1}(x))})\\&=\exp(\frac{\sec(x)}{\operatorname{gd}^{-1}(x)})\end{align}$$

$$ \operatorname{li}(x)= \int_{0}^{x}\frac{1}{\ln(t)}dt$$なので、
$$\begin{align}\operatorname{li}^*(x)&=\exp({\frac{d}{dx}\ln(\operatorname{li}(x))})\\&=\exp(\frac{1/\ln(x)}{\operatorname{li}(x)})\\&=\exp(\frac{1}{\ln(x)\operatorname{li}(x)})\end{align}$$

$$\int_{}^{}\frac{e^x}{x}dx=\operatorname{Ei}(x)+C$$なので、
$$\begin{align}\operatorname{Ei}^*(x)&=\exp({\frac{d}{dx}\ln(\operatorname{Ei}(x))})\\&=\exp(\frac{e^x/x}{\operatorname{Ei}(x)})\\&=\exp(\frac{e^x}{x\operatorname{Ei}(x)})\end{align}$$

$$ \operatorname{Si}(x)= \int_{0}^{x}\frac{\sin(t)}{t}dt$$なので、
$$\begin{align}\operatorname{Si}^*(x)&=\exp({\frac{d}{dx}\ln(\operatorname{Si}(x))})\\&=\exp(\frac{\sin(x)/x}{\operatorname{Si}(x)})\\&=\exp(\frac{\sin(x)}{x\operatorname{Si}(x)})\end{align}$$

$$ \operatorname{Ci}(x)= \int_{0}^{x}\frac{\cos(t)}{t}dt$$なので、
$$\begin{align}\operatorname{Ci}^*(x)&=\exp({\frac{d}{dx}\ln(\operatorname{Ci}(x))})\\&=\exp(\frac{\cos(x)/x}{\operatorname{Ci}(x)})\\&=\exp(\frac{\cos(x)}{x\operatorname{Ci}(x)})\end{align}$$

$$ \operatorname{S}(x)= \int_{0}^{x}\sin(t^2)dt$$なので、
$$\begin{align}\operatorname{S}^*(x)&=\exp({\frac{d}{dx}\ln(\operatorname{S}(x))})\\&=\exp(\frac{\sin(x^2)}{\operatorname{S}(x)})\end{align}$$

$$ \operatorname{C}(x)= \int_{0}^{x}\cos(t^2)dt$$なので、
$$\begin{align}\operatorname{C}^*(x)&=\exp({\frac{d}{dx}\ln(\operatorname{C}(x))})\\&=\exp(\frac{\cos(x^2)}{\operatorname{C}(x)})\end{align}$$

$$ \erf(x)=\frac{2}{\sqrt{\pi}}\int_{0}^{x}e^{-t^2}dt$$なので、
$$\begin{align}\erf^*(x)&=\exp({\frac{d}{dx}\ln(\erf(x))})\\&=\exp(\frac{2e^{-x^2}/\sqrt{\pi}}{\erf(x)})\\&=\exp(\frac{2e^{-x^2}}{\sqrt{\pi}\erf(x)})\end{align}$$

$$\begin{align}\operatorname{erfc}(x)&=1-\erf(x)\\&=1-\frac{2}{\sqrt{\pi}}\int_{0}^{x}e^{-t^2}dt\end{align}$$なので、
$$\begin{align}\operatorname{erfc}^*(x)&=\exp({\frac{d}{dx}\ln(\operatorname{erfc}(x))})\\&=\exp(\frac{-2e^{-x^2}/\sqrt{\pi}}{\operatorname{erfc}(x)})\\&=\exp(-\frac{2e^{-x^2}}{\sqrt{\pi}\operatorname{erfc}(x)})\end{align}$$

$$\operatorname{erfcx}(x)=e^{x^2}\operatorname{erfc}(x)$$なので、
$$\begin{align}\operatorname{erfcx}^*(x)&=\exp({\frac{d}{dx}\ln(\operatorname{erfcx}(x))})\\&=\exp(\frac{2xe^{x^2}\operatorname{erfc}(x)+e^{x^2}(-2e^{-x^2}/\sqrt{\pi})}{\operatorname{erfcx}(x)})\\&=\exp(\frac{2x\operatorname{erfcx}(x)-2/\sqrt{\pi}}{\operatorname{erfcx}(x)})\\&=\exp(2x-\frac{2}{\sqrt{\pi}\operatorname{erfcx}(x)})\end{align}$$

$$\begin{align}\operatorname{erfi}(x)&=-i\erf(ix)\\&=\frac{2}{\sqrt{\pi}}\int_{0}^{x}e^{t^2}dt\end{align}$$なので、
$$\begin{align}\operatorname{erfi}^*(x)&=\exp({\frac{d}{dx}\ln(\operatorname{erfi}(x))})\\&=\exp(\frac{2e^{x^2}/\sqrt{\pi}}{\operatorname{erfi}(x)})\\&=\exp(\frac{2e^{x^2}}{\sqrt{\pi}\operatorname{erfi}(x)})\end{align}$$

$$\operatorname{F}(x)=e^{-x^2}\int_{0}^{x}e^{t^2}dt$$なので、
$$\begin{align}\operatorname{F}^*(x)&=\exp({\frac{d}{dx}\ln(\operatorname{F}(x))})\\&=\exp(\frac{-2xe^{-x^2}\int_{0}^{x}e^{t^2}dt+e^{-x^2}e^{x^2}}{\operatorname{F}(x)})\\&=\exp(\frac{-2x\operatorname{F}(x)+1}{\operatorname{F}(x)})\\&=\exp(\frac{1}{\operatorname{F}(x)}-2x)\end{align}$$

$$\begin{align}\Gamma^*(x)&=\exp(\frac{d}{dx}\ln{(\Gamma (x))})\\&=\exp(\frac{\Gamma^{\prime}(x)}{\Gamma(x)})\\&=e^{\operatorname{\psi}(x)}\end{align}$$

$$\begin{align}\operatorname{\psi}^*(x)&=\exp(\frac{d}{dx}\ln{(\operatorname{\psi} (x))})\\&=\exp(\frac{\operatorname{\psi}^{\prime}(x)}{\operatorname{\psi}(x)})\\&=e^{\operatorname{\psi}^{(1)}(x)}\end{align}$$

$$\begin{align}{\operatorname{\psi}^{(n)}}^*(x)&=\exp(\frac{d}{dx}\ln{(\operatorname{\psi}^{(n)}(x))})\\&=\exp(\frac{{\operatorname{\psi}^{(n)}}^{\prime}(x)}{\operatorname{\psi}^{(n)}}{(x)})\\&=e^{\operatorname{\psi}^{(n+1)}(x)}\end{align}$$

$$\operatorname{B}(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$$なので、
$$\begin{align}\operatorname{B}_{x}^*(x,y)&=\exp(\frac{\partial}{\partial x}\ln{(\operatorname{B}(x,y) )})\\&=\exp({\frac{\partial}{\partial x}(\ln(\Gamma(x))+\ln(\Gamma(y))-\ln(\Gamma(x+y))})\\&=\exp(\operatorname{\psi}(x)-\operatorname{\psi}(x+y))\end{align}$$

$$\operatorname{B}(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$$なので、
$$\begin{align}\operatorname{B}_{xy}^{**}(x,y)&=\exp(\frac{{\partial}^2}{\partial x\partial y}\ln{(\operatorname{B}(x,y) )})\\&=\exp({\frac{{\partial}^2}{\partial x\partial y}(\ln(\Gamma(x))+\ln(\Gamma(y))-\ln(\Gamma(x+y)))})\\&=\exp({\frac{\partial}{\partial y}\operatorname{\psi}(x)-\operatorname{\psi}(x+y)})\\&=\exp(-\operatorname{\psi}^{(1)}{(x+y)})\end{align}$$

$$\begin{align}\operatorname{W}^*(x)&=\exp(\frac{d}{dx}\ln{(\operatorname{W} (x))})\\&=\exp(\frac{\operatorname{W}^{\prime}(x)}{\operatorname{W}(x)})\end{align}$$
ここで、
$$\begin{align}{\operatorname{W}^{\prime}(x)}&=\frac{d}{dx}\operatorname{W}(x)\\&=\frac{1}{dx/d\operatorname{W}}\end{align}$$
$x=\operatorname{W}(x)e^{\operatorname{W}(x)}$なので、
$$\begin{align}\frac{1}{dx/d\operatorname{W}}&=\frac{1}{d\operatorname{W}(x)e^{\operatorname{W}(x)}/d\operatorname{W}}\\&=\frac{1}{e^{\operatorname{W}(x)}(1+\operatorname{W}(x))}\\&=\frac{\operatorname{W}(x)}{x(1+\operatorname{W}(x))}\end{align}$$
よって、$$\operatorname{W}^*(x)=\exp(\frac{1}{x(1+\operatorname{W}(x))})$$

$$\operatorname{F}(x,k)=\int_{0}^{x}\frac{dt}{\sqrt{(1-t^2)(1-k^2t^2)}}$$なので、
$$\begin{align}\operatorname{F}_{x}^*(x,k)&=\exp(\frac{\partial}{\partial x}\ln{(\operatorname{F}(x,k))})\\&=\exp(\frac{1}{\operatorname{F}(x,k)\sqrt{(1-x^2)(1-k^2 x^2)}})\end{align}$$

$$\operatorname{F}(x,k)=\int_{0}^{x}\frac{dt}{\sqrt{(1-t^2)(1-k^2t^2)}}$$なので、
$$\begin{align}\operatorname{F}_{k}^*(x,k)&=\exp(\frac{\partial}{\partial k}\ln{(\operatorname{F}(x,k))})\end{align}$$
ここでライプニッツの積分法則を使いたいが、$\operatorname{F}(x,k)$の被積分関数は三変数関数である。そのため、新たにライプニッツの積分公式を拡張する必要がある。
例えば、
$$F(x,k)=\int_{a}^{x}f(x,k,t)dt$$
とする。
偏微分の定義より、
$$\begin{align}\frac{\partial}{\partial k}F(x,k)&= \lim_{\varDelta k \to 0} \frac{F(x,k+\varDelta k )-F(x,k)}{\varDelta k}\\&=\lim_{\varDelta k \to 0}\frac{\int_{a}^{x}f(x,k+\varDelta k,t)dt-\int_{a}^{x}f(x,k,t)dt}{\varDelta k}\\&=\lim_{\varDelta k \to 0}\int_{a}^{x}\frac{f(x,k+\varDelta k,t)dt-f(x,k,t)dt}{\varDelta k}\end{align}$$
ここで、差分商が一様収束することを証明する。
$$g_{\varDelta k}(t)=\frac{f(x,k+\varDelta k,t)-f(x,k,t)}{\varDelta k}$$とする。
$\varDelta k \to 0$のときに、
$$g(t)=\frac{\partial}{\partial k}f(x,k,t)$$が$t \in I_t=[a,x]$で一様収束することを示す。
平均値の定理より、
$$f(x,k+\varDelta k,t)-f(x,k,t)=\frac{\partial}{\partial k}f(x,k+\theta \varDelta k,t)\varDelta k$$
ただし$\theta$は$0<\theta<1$である。このとき、$I_k=[k-\delta_0,k+\delta_0]$に特異点が現れないような、連続性が保たれる十分に小さい$\delta_0$を設定する。
$I_t,I_k$はともに有界閉区間なので、$D=I_t \times I_k$もコンパクト集合となる。
$\frac{\partial}{\partial k}f(x,k,t)$は領域$D$で連続であるため、コンパクト集合上の連続関数の性質より、一様連続である。
一様連続の定義より、任意の$\varepsilon>0$に対し、ある$\delta<0$が存在して、$|\kappa_1-\kappa_2|<\delta$を満たす全ての$\kappa_1,\kappa_2 \in I_k$と全ての$t \in I_t$に対し、
$$\left|\frac{\partial}{\partial k}f(x,\kappa_1,t)-\frac{\partial}{\partial k}f(x,\kappa_2,t)\right|<\varepsilon$$が成り立つ。
ここで、$\kappa_1=k+\theta\varDelta k,\kappa_2=k$とする。つまり、
$$\left|\frac{\partial}{\partial k}f(x,k+\theta\varDelta k,t)-\frac{\partial}{\partial k}f(x,k,t)\right|<\varepsilon$$である。
$$f(x,k+\varDelta k,t)-f(x,k,t)=\frac{\partial}{\partial k}f(x,k+\theta \varDelta k,t)\varDelta k$$なので、
$$\left|g_{\varDelta k}(t)-g(t)\right|<\varepsilon$$
全ての$t$で成り立つので、上限を取り、
$$\sup_{t\in[a,x]}\left|g_{\varDelta k}(t)-g(t)\right|\leq\varepsilon$$
も成り立つ。$\varepsilon>0$は任意に小さくできるので、$$\lim_{\varDelta k\to 0}\sup_{t\in[a,x]}\left|g_{\varDelta k}(t)-g(t)\right|=0$$
よって、一様収束を証明することができた。
$$\left|\int_{a}^{x}g_{\varDelta k}(t)dt-\int_{a}^{x}g(t)dt\right|$$を評価する。積分の線形性より、
$$=\left|\int_{a}^{x}(g_{\varDelta k}(t)-g(t))dt\right|$$
三角不等式より、
$$\leq\int_{a}^{x}\left|g_{\varDelta k}(t)-g(t)\right|dt$$
$$\left|g_{\varDelta k}(t)-g(t)\right|\leq\sup_{t \in[a,x]}\left|g_{\varDelta k}(t)-g(t)\right|$$なので、
$$\leq\int_{a}^{x}\sup_{t \in[a,x]}\left|g_{\varDelta k}(t)-g(t)\right|dt$$
被積分関数は定数なので、
$$=(x-a)\sup_{t \in[a,x]}\left|g_{\varDelta k}(t)-g(t)\right|$$
$$\lim_{\varDelta k\to 0}\sup_{t\in[a,x]}\left|g_{\varDelta k}(t)-g(t)\right|=0$$なので、
$$\lim_{\varDelta k\to 0}(x-a)\sup_{t \in[a,x]}\left|g_{\varDelta k}(t)-g(t)\right|=0$$
$$\left|\int_{a}^{x}g_{\varDelta k}(t)dt-\int_{a}^{x}g(t)dt\right|\leq(x-a)\sup_{t\in[a,x]}\left|g_{\varDelta k}(t)-g(t)\right|$$
を既に示したので、
$$\lim_{\varDelta k\to 0}\left(\int_{a}^{x}g_{\varDelta k}(t)dt-\int_{a}^{x}g(t)dt\right)=0$$
$$\lim_{\varDelta k\to 0}\int_{a}^{x}g_{\varDelta k}(t)dt=\int_{a}^{x}g(t)dt$$
$$\lim_{\varDelta k\to 0}\int_{a}^{x}g_{\varDelta k}(t)dt=\int_{a}^{x}\lim_{\varDelta k\to 0}g_{\varDelta k}dt$$
積分と極限が交換できた。あとは偏微分を含んだ式に戻して、
$$\begin{align}\lim_{\varDelta k \to 0}\int_{a}^{x}\frac{f(x,k+\varDelta k,t)dt-f(x,k,t)dt}{\varDelta k}&=\int_{a}^{x}\lim_{\varDelta k \to 0}\frac{f(x,k+\varDelta k,t)dt-f(x,k,t)dt}{\varDelta k}\\&=\int_{a}^{x}\frac{\partial}{\partial k}f(x,k,t)dt\end{align}$$
よって、$$\frac{\partial}{\partial k}\int_{a}^{x}f(x,k,t)=\int_{a}^{x}\frac{\partial}{\partial k}f(x,k,t)dt $$が成り立ち、二変数関数にライプニッツの積分公式が拡張された。
ここで、$0\leq |t|\leq|x|<1,|k|<1$とし、$\operatorname{F}(x,k)$の被積分関数$$\frac{1}{\sqrt{(1-t^2)(1-k^2t^2)}}$$はこの範囲で連続である。
また、多項式$$(1-t^2)(1-k^2t^2)$$はこの範囲で微分可能であり、その正の平方根、さらにその逆数もこの範囲で微分可能である。
したがって、平均値の定理での条件、あるいはその他全ての条件に当てはまっているので、拡張されたライプニッツの積分公式が使える。
よって、
$$\begin{align}\operatorname{F}_{k}^*(x,k)&=\exp(\frac{\partial}{\partial k}\ln{(\operatorname{F}(x,k))})\\&=\exp(\frac{(\partial/\partial k)\operatorname{F}(x,k)}{\operatorname{F}(x,k)})\\&=\exp(\frac{(\partial/\partial k)\int_{0}^{x}dt/\sqrt{(1-t^2)(1-k^2t^2)}}{\operatorname{F}(x,k)})\\&=\exp(\frac{\int_{0}^{x}(\partial/\partial k)dt/\sqrt{(1-t^2)(1-k^2t^2)}}{\operatorname{F}(x,k)})\\&=\exp(\frac{\int_{0}^{x}-(1-k^2t^2)^{-3/2}(-2kt^2)dt/2\sqrt{(1-t^2)}}{\operatorname{F}(x,k)})\\&=\exp(\frac{1}{\operatorname{F}(x,k)}\int_{0}^{x}\frac{kt^2}{\sqrt{(1-t^2)}(1-k^2t^2)^{3/2}}dt)\end{align}$$

$$\operatorname{E}(x,k)=\int_{0}^{x}\sqrt{\frac{1-k^2t^2}{1-t^2}}dt$$なので、
$$\begin{align}\operatorname{E}_{x}^*(x,k)&=\exp(\frac{\partial}{\partial x}\ln{(\operatorname{E}(x,k))})\\&=\exp(\frac{1}{\operatorname{E}(x,k)}\sqrt{\frac{1-k^2x^2}{1-x^2}})\end{align}$$

$$\operatorname{E}(x,k)=\int_{0}^{x}\sqrt{\frac{1-k^2t^2}{1-t^2}}dt$$であり、被積分関数は$0\leq |t|\leq|x|<1,|k|<1$で連続、微分可能なので、拡張されたライプニッツの積分公式を使って、
$$\begin{align}\operatorname{E}_{k}^*(x,k)&=\exp(\frac{\partial}{\partial k}\ln{(\operatorname{E}(x,k))})\\&=\exp(\frac{(\partial/\partial k)\operatorname{E}(x,k)}{\operatorname{E}(x,k)})\\&=\exp(\frac{(\partial/\partial k)\int_{0}^{x}\sqrt{(1-k^2t^2)/(1-t^2)}dt}{\operatorname{E}(x,k)})\\&=\exp(\frac{\int_{0}^{x}(\partial/\partial k)\sqrt{(1-k^2t^2)/(1-t^2)}dt}{\operatorname{E}(x,k)})\\&=\exp(\frac{\int_{0}^{x}(1/2)(1-k^2t^2)^{-1/2}(-2kt^2)/\sqrt{1-t^2}dt}{\operatorname{E}(x,k)})\\&=\exp(-\frac{1}{\operatorname{E}(x,k)}\int_{0}^{x}\frac{kt^2}{\sqrt{(1-t^2)(1-k^2t^2)}}dt)\end{align}$$

$$\Pi(a;x,k)=\int_{0}^{x}\frac{dt}{(1-at^2)\sqrt{(1-t^2)(1-k^2t^2)}}$$なので、
$$\begin{align}\Pi_{x}^*(a;x,k)&=\exp(\frac{\partial}{\partial x}\ln{(\Pi(a;x,k))})\\&=\exp(\frac{1}{\Pi(a;x,k)(1-ax^2)\sqrt{(1-x^2)(1-k^2x^2)}})\end{align}$$

$$\Pi(a;x,k)=\int_{0}^{x}\frac{dt}{(1-at^2)\sqrt{(1-t^2)(1-k^2t^2)}}$$である。
ここで、ライプニッツの積分公式を使いたいが、対象の関数は三変数関数である。しかし、二変数関数に拡張されたライプニッツの積分公式について、三変数関数にも同様の議論が成り立つので、ある範囲で条件を満たせば、三変数関数にも適用可能である。
被積分関数は$0\leq |t|\leq|x|<1,|k|<1,|a|<\frac{1}{x^2}$で連続、微分可能なので、拡張されたライプニッツの積分公式を使って、
$$\begin{align}\Pi_{k}^*(a;x,k)&=\exp(\frac{\partial}{\partial k}\ln{(\Pi(a;x,k))})\\&=\exp(\frac{(\partial/\partial k)\Pi(a;x,k)}{\Pi(a;x,k)})\\&=\exp(\frac{(\partial/\partial k)\int_{0}^{x}dt/((1-at^2)\sqrt{(1-t^2)(1-k^2t^2)})}{\Pi(a;x,k)})\\&=\exp(\frac{\int_{0}^{x}(\partial/\partial k)(dt/((1-at^2)\sqrt{(1-t^2)(1-k^2t^2)}))}{\Pi(a;x,k)})\\&=\exp(\frac{\int_{0}^{x}(-2kt^2)(-1/2)(1-k^2t^2)^{-3/2}/((1-at^2)\sqrt{(1-t^2)})dt}{\Pi(a;x,k)})\\&=\exp(\frac{1}{\Pi(a;x,k)}\int_{0}^{x}\frac{kt^2}{(1-at^2)\sqrt{1-t^2}(1-k^2t^2)^{3/2}}dt)\end{align}$$

$$\Pi(a;x,k)=\int_{0}^{x}\frac{dt}{(1-at^2)\sqrt{(1-t^2)(1-k^2t^2)}}$$である。
被積分関数は$0\leq |t|\leq|x|<1,|k|<1,|a|<\frac{1}{x^2}$で連続、微分可能より、平均値の定理の条件およびその他の全ての条件が成り立つので、拡張されたライプニッツの積分公式を使って、
$$\begin{align}\Pi_{a}^*(a;x,k)&=\exp(\frac{\partial}{\partial a}\ln{(\Pi(a;x,k))})\\&=\exp(\frac{(\partial/\partial a)\Pi(a;x,k)}{\Pi(a;x,k)})\\&=\exp(\frac{(\partial/\partial a)\int_{0}^{x}dt/((1-at^2)\sqrt{(1-t^2)(1-k^2t^2)})}{\Pi(a;x,k)})\\&=\exp(\frac{\int_{0}^{x}(\partial/\partial a)(dt/((1-at^2)\sqrt{(1-t^2)(1-k^2t^2)}))}{\Pi(a;x,k)})\\&=\exp(\frac{\int_{0}^{x}(t^2((1-at^2)^{-2}/\sqrt{(1-t^2)(1-k^2t^2)}))dt}{\Pi(a;x,k)})\\&=\exp(\frac{1}{\Pi(a;x,k)}\int_{0}^{x}\frac{t^2}{(1-at^2)^2\sqrt{(1-t^2)(1-k^2t^2)}}dt)\end{align}$$

$$\operatorname{am}_u'(u,k)=\operatorname{dn}(u,k)$$なので、
$$\begin{align}\operatorname{am}_u^*(u,k)&=\exp(\frac{\partial}{\partial u}\ln{(\operatorname{am} (u,k))})\\&=\exp(\frac{\operatorname{dn}(u,k)}{\operatorname{am}(u,k)})\end{align}$$

$$\begin{align}\operatorname{sn}_u^*(u,k)&=\exp(\frac{\partial}{\partial u}\ln{(\operatorname{sn} (u,k))})\\&=\exp(\frac{\operatorname{cn}(u,k)\operatorname{dn}(u,k)}{\operatorname{sn}(u,k)})\end{align}$$

$$\begin{align}\operatorname{cn}_u^*(u,k)&=\exp(\frac{\partial}{\partial u}\ln{(\operatorname{cn} (u,k))})\\&=\exp(-\frac{\operatorname{sn}(u,k)\operatorname{dn}(u,k)}{\operatorname{cn}(u,k)})\end{align}$$

$$\begin{align}\operatorname{dn}_u^*(u,k)&=\exp(\frac{\partial}{\partial u}\ln{(\operatorname{dn} (u,k))})\\&=\exp(-\frac{k^2\operatorname{sn}(u,k)\operatorname{cn}(u,k)}{\operatorname{dn}(u,k)})\end{align}$$

$$\operatorname{ns}(u,k)=\frac{1}{\operatorname{sn}(u,k)}$$なので、
$$\begin{align}\operatorname{ns}_u^*(u,k)&=\exp(\frac{\partial}{\partial u}\ln{(\operatorname{ns} (u,k))})\\&=\exp(\frac{1}{\operatorname{ns}(u,k)}\left({\frac{\partial1}{\partial u}-\frac{\partial \operatorname{sn}(u,k)}{\partial u}}\right))\\&=\exp(-\frac{\operatorname{cn}(u,k)\operatorname{dn}(u,k)}{\operatorname{sn}(u,k)})\end{align}$$

$$\operatorname{nc}(u,k)=\frac{1}{\operatorname{cn}(u,k)}$$なので、
$$\begin{align}\operatorname{nc}_u^*(u,k)&=\exp(\frac{\partial}{\partial u}\ln{(\operatorname{nc} (u,k))})\\&=\exp(\frac{1}{\operatorname{nc}(u,k)}\left(\frac{\partial1}{\partial u}-\frac{\partial \operatorname{cn}(u,k)}{\partial u}\right))\\&=\exp(\frac{\operatorname{sn}(u,k)\operatorname{dn}(u,k)}{\operatorname{cn}(u,k)})\end{align}$$

$$\operatorname{nd}(u,k)=\frac{1}{\operatorname{dn}(u,k)}$$なので、
$$\begin{align}\operatorname{nd}_u^*(u,k)&=\exp(\frac{\partial}{\partial u}\ln{(\operatorname{dn} (u,k))})\\&=\exp(\frac{1}{\operatorname{nd(u,k)}}\left(\frac{\partial1}{\partial u}-\frac{\partial \operatorname{dn}(u,k)}{\partial u}\right))\\&=\exp(\frac{k^2\operatorname{sn}(u,k)\operatorname{cn}(u,k)}{\operatorname{dn}(u,k)})\end{align}$$

$$\operatorname{sc}(u,k)=\frac{\operatorname{sn}(u,k)}{\operatorname{cn}(u,k)}$$なので、
$$\begin{align}\operatorname{sc}_u^*(u,k)&=\exp(\frac{\partial}{\partial u}\ln{(\operatorname{sc} (u,k))})\\&=\exp(\frac{1}{\operatorname{sc}(u,k)}\left(\frac{\partial\operatorname{sn}(u,k)}{\partial u}-\frac{\partial \operatorname{cn}(u,k)}{\partial u}\right))\\&=\exp(\frac{\operatorname{dn}(u,k)}{\operatorname{sn}(u,k)\operatorname{cn}(u,k)})\end{align}$$

$$\operatorname{cs}(u,k)=\frac{\operatorname{cn}(u,k)}{\operatorname{sn}(u,k)}$$なので、
$$\begin{align}\operatorname{cs}_u^*(u,k)&=\exp(\frac{\partial}{\partial u}\ln{(\operatorname{cs} (u,k))})\\&=\exp(\frac{1}{\operatorname{cs}(u,k)}\left(\frac{\partial\operatorname{sn}(u,k)}{\partial u}-\frac{\partial \operatorname{cn}(u,k)}{\partial u}\right))\\&=\exp(-\frac{\operatorname{dn}(u,k)}{\operatorname{sn}(u,k)\operatorname{cn}(u,k)})\end{align}$$

$$\operatorname{sd}(u,k)=\frac{\operatorname{sn}(u,k)}{\operatorname{dn}(u,k)}$$なので、
$$\begin{align}\operatorname{sd}_u^*(u,k)&=\exp(\frac{\partial}{\partial u}\ln{(\operatorname{sd} (u,k))})\\&=\exp(\frac{1}{\operatorname{sd}(u,k)}\left(\frac{\partial\operatorname{sn}(u,k)}{\partial u}-\frac{\partial \operatorname{dn}(u,k)}{\partial u}\right))\\&=\exp(\frac{(1-k^2)\operatorname{cn}(u,k)}{\operatorname{sn}(u,k)\operatorname{dn}(u,k)})\end{align}$$

$$\operatorname{ds}(u,k)=\frac{\operatorname{dn}(u,k)}{\operatorname{sn}(u,k)}$$なので、
$$\begin{align}\operatorname{ds}_u^*(u,k)&=\exp(\frac{\partial}{\partial u}\ln{(\operatorname{ds} (u,k))})\\&=\exp(\frac{1}{\operatorname{ds}(u,k)}\left(\frac{\partial\operatorname{dn}(u,k)}{\partial u}-\frac{\partial \operatorname{sn}(u,k)}{\partial u}\right))\\&=\exp(-\frac{(1-k^2)\operatorname{cn}(u,k)}{\operatorname{sn}(u,k)\operatorname{dn}(u,k)})\end{align}$$

$$\operatorname{cd}(u,k)=\frac{\operatorname{cn}(u,k)}{\operatorname{dn}(u,k)}$$なので、
$$\begin{align}\operatorname{cd}_u^*(u,k)&=\exp(\frac{\partial}{\partial u}\ln{(\operatorname{cd} (u,k))})\\&=\exp(\frac{1}{\operatorname{cd}(u,k)}\left(\frac{\partial\operatorname{cn}(u,k)}{\partial u}-\frac{\partial \operatorname{dn}(u,k)}{\partial u}\right))\\&=\exp(-\frac{(1-k^2)\operatorname{sc}(u,k)}{\operatorname{cn}(u,k)\operatorname{dn}(u,k)})\end{align}$$

$$\operatorname{dc}(u,k)=\frac{\operatorname{dn}(u,k)}{\operatorname{cn}(u,k)}$$なので、
$$\begin{align}\operatorname{dc}_u^*(u,k)&=\exp(\frac{\partial}{\partial u}\ln{(\operatorname{dc} (u,k))})\\&=\exp(\frac{1}{\operatorname{dc}(u,k)}\left(\frac{\partial\operatorname{dn}(u,k)}{\partial u}-\frac{\partial \operatorname{cn}(u,k)}{\partial u}\right))\\&=\exp(\frac{(1-k^2)\operatorname{sc}(u,k)}{\operatorname{cn}(u,k)\operatorname{dn}(u,k)})\end{align}$$

$$\begin{align}\operatorname{arcsn}(x,k)&=\operatorname{F}(x,k)\\&=\int_{0}^{x}\frac{dt}{\sqrt{(1-t^2)(1-k^2t^2)}}\end{align}$$なので、
$$\begin{align}\operatorname{arcsn}_{x}^*(x,k)&=\exp(\frac{\partial}{\partial x}\ln{(\operatorname{arcsn}(x,k))})\\&=\exp(\frac{1}{\operatorname{arcsn}(x,k)\sqrt{(1-x^2)(1-k^2 x^2)}})\end{align}$$

$$\operatorname{cn}'(u,k)=-\operatorname{sn}(u,k)\operatorname{dn}(u,k),\operatorname{sn}(u,k)=\sqrt{1-x^2},\operatorname{dn}(u,k)=\sqrt{1-k^2+k^2x^2}$$なので、
$$\begin{align}\operatorname{arccn}_{x}^*(x,k)&=\exp(\frac{\partial}{\partial x}\ln{(\operatorname{arccn}(x,k))})\\&=\exp(-\frac{1}{\operatorname{arccn}(x,k)\sqrt{(1-x^2)(1-k^2+k^2x^2)}})\end{align}$$

$$\operatorname{dn}'(u,k)=-k^2\operatorname{sn}(u,k)\operatorname{cn}(u,k),\operatorname{sn}(u,k)=\frac{\sqrt{1-x^2}}{k},\operatorname{dn}(u,k)=\frac{\sqrt{k^2-1+x^2}}{k}$$なので、
$$\begin{align}\operatorname{arcdn}_{x}^*(x,k)&=\exp(\frac{\partial}{\partial x}\ln{(\operatorname{arcdn}(x,k))})\\&=\exp(-\frac{1}{\operatorname{arcdn}(x,k)\sqrt{(1-x^2)(k^2-1+x^2)}})\end{align}$$

$$\operatorname{ns}'(u,k)=-\frac{\operatorname{cn}(u,k)\operatorname{dn}(u,k)}{\operatorname{sn}^2(u,k)},\operatorname{cn}(u,k)=\frac{\sqrt{x^2-1}}{x},\operatorname{dn}(u,k)=\frac{\sqrt{x^2-k^2}}{x},\operatorname{sn}(u,k)=\frac{1}{x}$$なので、
$$\begin{align}\operatorname{arcns}_{x}^*(x,k)&=\exp(\frac{\partial}{\partial x}\ln{(\operatorname{arcns}(x,k))})\\&=\exp(-\frac{1}{\operatorname{arcns}(x,k)\sqrt{(x^2-1)(x^2-k^2)}})\end{align}$$

$$\operatorname{nc}'(u,k)=\frac{\operatorname{sn}(u,k)\operatorname{dn}(u,k)}{\operatorname{cn}^2(u,k)},\operatorname{sn}(u,k)=\frac{\sqrt{x^2-1}}{x},\operatorname{dn}(u,k)=\frac{\sqrt{(1-k^2)x^2+k^2}}{x},\operatorname{cn}(u,k)=\frac{1}{x}$$なので、
$$\begin{align}\operatorname{arcnc}_{x}^*(x,k)&=\exp(\frac{\partial}{\partial x}\ln{(\operatorname{arcnc}(x,k))})\\&=\exp(\frac{1}{\operatorname{arcnc}(x,k)\sqrt{(x^2-1)((1-k^2)x^2+k^2)}})\end{align}$$

$$\operatorname{nd}'(u,k)=\frac{k^2\operatorname{sn}(u,k)\operatorname{cn}(u,k)}{\operatorname{dn}^2(u,k)},\operatorname{sn}(u,k)=\frac{\sqrt{x^2-1}}{x},\operatorname{cn}(u,k)=\frac{\sqrt{k^2x^2+1-k^2}}{kx},\operatorname{dn}(u,k)=\frac{1}{x}$$なので、
$$\begin{align}\operatorname{arcnd}_{x}^*(x,k)&=\exp(\frac{\partial}{\partial x}\ln{(\operatorname{arcnd}(x,k))})\\&=\exp(\frac{1}{k\operatorname{arcnd}(x,k)\sqrt{(x^2-1)(k^2x^2+1-k^2)}})\end{align}$$

$$\operatorname{sc}'(u,k)=\frac{\operatorname{dn}(u,k)}{\operatorname{cn}^2(u,k)},\operatorname{dn}(u,k)=\frac{\sqrt{1+(1-k^2)x^2}}{\sqrt{1+x^2}},\operatorname{cn}(u,k)=\frac{1}{\sqrt{1+x^2}}$$なので、
$$\begin{align}\operatorname{arcsc}_{x}^*(x,k)&=\exp(\frac{\partial}{\partial x}\ln{(\operatorname{arcsc}(x,k))})\\&=\exp(\frac{1}{\operatorname{arcsc}(x,k)\sqrt{(1+x^2)(1+(1-k^2)x^2)}})\end{align}$$

$$\operatorname{cs}'(u,k)=-\frac{k^2\operatorname{dn}(u,k)}{\operatorname{sn}^2(u,k)},\operatorname{dn}(u,k)=\frac{\sqrt{1-k^2+x^2}}{1+x^2},\operatorname{sn}(u,k)=\frac{1}{\sqrt{1+x^2}}$$なので、
$$\begin{align}\operatorname{arccs}_{x}^*(x,k)&=\exp(\frac{\partial}{\partial x}\ln{(\operatorname{arccs}(x,k))})\\&=\exp(-\frac{1}{\operatorname{arccs}(x,k)\sqrt{(1+x^2)(1-k^2+x^2)}})\end{align}$$

$$\operatorname{sd}'(u,k)=\frac{\operatorname{cn}(u,k)}{\operatorname{dn}^2(u,k)},\operatorname{cn}(u,k)=\frac{\sqrt{1+(1-k^2)x^2}}{\sqrt{1+k^2x^2}},\operatorname{dn}(u,k)=\frac{1}{\sqrt{1+k^2x^2}}$$なので、
$$\begin{align}\operatorname{arcsd}_{x}^*(x,k)&=\exp(\frac{\partial}{\partial x}\ln{(\operatorname{arcsd}(x,k))})\\&=\exp(\frac{1}{\operatorname{arcsd}(x,k)\sqrt{(1+k^2x^2)(1+(1-k^2)x^2)}})\end{align}$$

$$\operatorname{ds}'(u,k)=-\frac{\operatorname{cn}(u,k)}{\operatorname{sn}^2(u,k)},\operatorname{cn}(u,k)=\frac{\sqrt{1-k^2+x^2}}{\sqrt{x^2+k^2}},\operatorname{sn}(u,k)=\frac{1}{\sqrt{x^2+k^2}}$$なので、
$$\begin{align}\operatorname{arcds}_{x}^*(x,k)&=\exp(\frac{\partial}{\partial x}\ln{(\operatorname{arcds}(x,k))})\\&=\exp(-\frac{1}{\operatorname{arcds}(x,k)\sqrt{(x^2+k^2)(1-k^2+x^2)}})\end{align}$$

$$\operatorname{cd}'(u,k)=-\frac{(1-k^2)\operatorname{sn}(u,k)}{\operatorname{dn}^2(u,k)},\operatorname{sn}(u,k)=\frac{\sqrt{1-x^2}}{\sqrt{1-k^2x^2}},\operatorname{dn}(u,k)=\frac{\sqrt{1-k^2}}{\sqrt{1-k^2x^2}}$$なので、
$$\begin{align}\operatorname{arccd}_{x}^*(x,k)&=\exp(\frac{\partial}{\partial x}\ln{(\operatorname{arccd}(x,k))})\\&=\exp(-\frac{1}{\operatorname{arccd}(x,k)\sqrt{(1-x^2)(1-k^2x^2)}})\end{align}$$

$$\operatorname{dc}'(u,k)=\frac{(1-k^2)\operatorname{sn}(u,k)}{\operatorname{cn}^2(u,k)},\operatorname{sn}(u,k)=\frac{\sqrt{x^2-1}}{\sqrt{x^2-k^2}},\operatorname{cn}(u,k)=\frac{\sqrt{1-k^2}}{\sqrt{x^2-k^2}}$$なので、
$$\begin{align}\operatorname{arcdc}_{x}^*(x,k)&=\exp(\frac{\partial}{\partial x}\ln{(\operatorname{arcdc}(x,k))})\\&=\exp(\frac{1}{\operatorname{arcdc}(x,k)\sqrt{(x^2-1)(x^2-k^2)}})\end{align}$$