# How to prove $\frac{d}{dx}\sinh x=\cosh x$ when $\sinh$ and $\cosh$ are defined by an integral?

Define $$\sinh$$ and $$\cosh$$ by $$x=\int_0^{\sinh x}\frac{dt}{\sqrt{t^2+1}},\, x\in\mathbb{R}$$ $$x=\int_1^{\cosh x}\frac{dt}{\sqrt{t^2-1}},\, x\ge 0$$ and define $$\cosh (-x)=\cosh x$$ for $$x\lt 0$$.

By the inverse function theorem, we have $$\frac{d}{dt}\sinh^{-1}t=\frac{1}{\sqrt{t^2+1}}\implies \frac{d}{dx}\sinh x=\sqrt{\sinh^2 x+1},$$ $$\frac{d}{dt}\cosh^{-1}t=\frac{1}{\sqrt{t^2-1}}\implies \frac{d}{dx}\cosh x=\operatorname{sgn}(x)\sqrt{\cosh^2 x-1}.$$

How can I show that $$\frac{d}{dx}\sinh x=\cosh x$$ and $$\frac{d}{dx}\cosh x=\sinh x$$?

I think I should use the identity $$\cosh^2 x-\sinh^2 x=1$$ but I don't know how to prove that identity from the integral definition. I also managed to prove $$\frac{d^2}{dx^2}\sinh x=\sinh x$$ but that doesn't seem to really help.

• Are you sure you don't mean to define $\sinh(x)$ to be the unique $a\in\mathbb{R}$ for which $x=\int_0^a...$? Feb 13, 2022 at 16:48
• @OliverHouse The definitions should be fine. $\sinh^{-1}u=\int_0^u \frac{dt}{\sqrt{t^2+1}}$ if and only if $x=\int_0^{\sinh x}\frac{dt}{\sqrt{t^2+1}}$. Feb 13, 2022 at 16:51

$$\newcommand{\d}{\,\mathrm{d}}$$Notice that if one makes the substitution (for $$u\ge0$$) $$u^2=t^2-1,\,u\d u=t\d t,\,\frac{u}{\sqrt{u^2+1}}\d u=\d t$$ then one finds:
\begin{align}\int_1^{\sqrt{\sinh^2x+1}}\frac{1}{\sqrt{t^2-1}}\d t&=\int_0^{\sinh x}\frac{u}{u}\frac{1}{\sqrt{u^2+1}}\d u\\&=\int_0^{\sinh x}\frac{1}{\sqrt{t^2+1}}\d t\\&=x\end{align}
Which implies that $$\cosh x=\sqrt{\sinh^2x+1}=\frac{\d}{\d x}\sinh x$$.