Derivada de funciones hiperbólicas inversas

$$\quad \frac{d}{dx}\argsinh u=\frac{1}{\sqrt{u^2+1}}\frac{du}{dx}$$ (13)

$$\quad \frac{d}{dx}\argcosh u=\frac{\pm 1}{\sqrt{u^2-1}}\frac{du}{dx}\\ + Si,\,\cosh^{-1}u>0, u>1,\quad - Si,\,\cosh^{-1}u<0, u>1$$ (14)

$$\quad \frac{d}{dx}\argtanh u=\frac{1}{1-u^{2}}\frac{du}{dx}\\ -1<u<1$$ (15)

$$\quad \frac{d}{dx}\argcoth u=\frac{1}{1-u^{2}}\frac{du}{dx}\\ u>1\, \textsl{o}\,u<-1$$ (16)

$$\quad \frac{d}{dx}\argsech u=\frac{\mp 1}{u\sqrt{1-u^2}}\frac{du}{dx}\\ - Si,\,\sech ^{-1}u>0,\quad + Si,\,\sech ^{-1}u<0, \quad 0<u<1$$ (17)

$$\quad \frac{d}{dx}\argcsch u=\frac{-1}{\vert u\vert\sqrt{1+u^2}}... ...}{dx}=\frac{\mp 1}{u\sqrt{1+u^2}}\frac{du}{dx}\\ - Si,\, u>0 \quad + Si, \,u>0$$ (18)