I have seen it proven directly using the $\frac{d}{dx}$ operator and making use of the exponentials, i.e $\frac{d}{dx} $ $\ sinh x $= $\frac{d}{dx}\frac{e^{x} +e^{-x}}{2} $
I was wondering if I could use the limits, I tried with $\ sinhx$ and reached the final form of :
$h\overset{lim}{\rightarrow} 0 \frac{e^{x}}{2} \frac{e^{h}-1}{h} + \frac{e^{-x}}{2} \frac{1- e^{-h}}{h}$ which I felt was a dead end. can it be evaluated, and if so, with steps.
$$\dfrac{d}{dx}\sinh(x)=\dfrac{d}{dx}\left(\dfrac{e^x+e^{-x}}{2}\right)=\lim_{h\to 0}\dfrac{\dfrac{e^{x+h}+e^{-(x+h)}}{2}-\dfrac{e^x+e^{-x}}{2}}{h}$$ $$=\lim_{h\to0}\dfrac{e^x}{2}\dfrac{\left(e^h-1\right)}{h}+\dfrac{e^{-x}}{2}\dfrac{\left(e^{-h}-1\right)}{h}$$ These limits are very well-known, but if you must, you could use L'Hopital's Rule or Taylor Series or other methods to see that $$\lim_{h\to0}\dfrac{\left(e^h-1\right)}{h}=1$$ and $$\lim_{h\to0}\dfrac{\left(e^{-h}-1\right)}{h}=-1$$ so that $$\lim_{h\to0}\dfrac{e^x}{2}\dfrac{\left(e^h-1\right)}{h}+\dfrac{e^{-x}}{2}\dfrac{\left(e^{-h}-1\right)}{h}=\dfrac{e^x-e^{-x}}{2}=\cosh(x)$$