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.
$\newcommand{\d}{\,\mathrm{d}}$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 $\cosh x=\sqrt{\sinh^2x+1}=\frac{\d}{\d x}\sinh x$.