Functions with Relatively High First Derivatives Compared To Second Derivatives I'm currently performing some experimental investigation and is has now become of interest to discover a function which on some interval has a derivative which is as relatively high as possible compared to its second derivative. I already know exponentials with bases $>>e$ are a good option, but are there any others that could be used.
 A: Let us make a little investigation over here. First and foremost, I assume that by "as relatively high as possible" it is meant that the following quantity - let's call it a difference - gets as large as possible:
$$D_f=\left|\frac{f'(x)-f''(x)}{f''(x)}\right|$$
So, let's check some special cases.
Firstly, let $f$ be a polynomial, let's say:
$$f(x)=f_0+f_1x+f_2x^2+\dots+f_nx^n\mbox{, where }f_i\in\mathbb{R}\ \forall i=1,2,\dots,n,\ f_n\neq0$$
So, we now calculate $D_f$, as follows:


*

*$$f'(x)=f_1+2f_2x+3f_3x^2+\dots+nf_nx^{n-1}$$

*$$f''(x)=2f_2+6f_3x+12f_4x^2+\dots+n(n-1)f_nx^{n-2}$$


So:
$$\begin{align*}D_f=&\left|\frac{f'(x)-f''(x)}{f''(x)}\right|=\\
=&\left|\frac{f_1+2f_2x+3f_3x^2+\dots+nf_nx^{n-1}-(2f_2+6f_3x+12f_4x^2+\dots+n(n-1)f_nx^{n-2})}{2f_2+6f_3x+12f_4x^2+\dots+n(n-1)f_nx^{n-2}}\right|=\\
=&\left|\frac{f_1-2f_2+(2f_2-6f_3)x+\dots+((n-1)f_{n-1}-n(n-1)f_n)x^{n-2}+nf_nx^{n-1}}{2f_2+6f_3x+12f_4x^2+\dots+n(n-1)f_{n}x^{n-2}}\right|=\\
=&\left|\frac{x^{n-1}\left(\frac{f_1-2f_2}{x^{n-1}}+\frac{2f_2-6f_3}{x^{n-2}}+\dots+\frac{(n-1)f_{n-1}-n(n-1)f_n}{x}+nf_n\right)}{x^{n-2}\left(\frac{2f_2}{x^{n-2}}+\frac{6f_3}{x^{n-3}}+\dots+n(n-1)f_n\right)}\right|=\\
=&|x|\cdot M(f,x,n)
\end{align*}$$
where
$$M(f,x,n)=\left|\frac{\frac{f_1-2f_2}{x^{n-1}}+\frac{2f_2-6f_3}{x^{n-2}}+\dots+\frac{(n-1)f_{n-1}-n(n-1)f_n}{x}+nf_n}{\frac{2f_2}{x^{n-2}}+\frac{6f_3}{x^{n-3}}+\dots+n(n-1)f_n}\right|$$
While $|x|\to+\infty$, we can safely assume - since $\lim\limits_{|x|\to+\infty}\frac{A}{x^p}=0$ for every $p>0$ - that 
$$M(f,x,n)\approx\left|\frac{nf_n}{n(n-1)f_n}\right|=\frac{1}{n-1}$$
So, by choosing an interval $[a,b]$ "away" from zero ($a>>1$) and a polynomial of rahter small degree (since $M\overset{n\to\infty}{\longrightarrow}0$ when $|x|>>1$, as shown above) one can get pretty nice examples of such functions. 
Note, of course, that we can "transport" the axis, e.g. $g(x)=f(x-a)$, to have a nicer interval, closer to zero.
Secondly, let us now assume that $f$ is in the form of 
$$f(x)=A\sin(ax)+B\cos(bx)$$
where $a,b,A,B\in\mathbb{R}$. Then, we have, as previously:


*

*$f'(x)=aA\cos(ax)-bB\sin(bx)$,

*$f''(x)=-a^2A\sin(ax)-b^2B\cos(bx)$


So:
$$\begin{align*}
D_f=&\left|\frac{aA\cos(ax)-bB\sin(bx)-(-a^2A\sin(ax)-b^2B\cos(bx))}{-a^2A\sin(ax)-b^2B\cos(bx)}\right|=\\
=&\left|\frac{aA(\cos(ax)+a\sin(ax))-bB(\sin(bx)-b\cos(bx))}{-a^2A\sin(ax)-b^2B\cos(bx)}\right|\leq\\
\leq&\frac{|aA(\cos(ax)+a\sin(ax))|+|bB(\sin(bx)-b\cos(bx))|}{|a^2A\sin(ax)+b^2B\sin(bx)|}\leq\\
\leq&\frac{|aA|(|\cos(ax)|+|a\sin(ax)|)+|bB|(|\sin(bx)+|b\cos(bx)|)}{|a^2A\sin(ax)+b^2B\cos(bx)|}\leq\\
\leq&\frac{|aA|(1+|a|)+|bB|(1+|b|)}{|a^2A\sin(ax)+b^2B\cos(bx)|}
\end{align*}$$
From this point, we have several options. One is to set $B=0$, so we have, after some simplifications:
$$D_f\leq\frac{1+|a|}{|a\sin(ax)|}$$
So, for $x\in\left(-\frac{\pi}{a},\frac{\pi}{a}\right)$ and for a "not-that-large" value of $a$, we see that the upper bound of $D_f$ is getting as large as we want.
Note that, in this case we have calculated an upper bound of $D_f$, so, it is unsafe to say that $D_f$ will behave in exactly the same way.
Lastly, every combination of the above might give interesting results, but I feel that there has been created a large wall of text, so I should keep it to this level.
Hope it helped! :)
