How is the gradient of exponential functions with different bases (in the included table) worked out? I am studying exponential functions at the moment, and this table was presented in my textbook to show that for exponential functions with increasing 'bases' the gradient of the function increases.

Okay I get that they get steeper as the base 'b' becomes larger, I understand that for base b=1 that the gradient is 0 as essentially what you have is a horizontal line; but for the others I'm trying to understand how that the figure multiplying the function was worked out? i.e. the 0.7 in the $y=2^x$ row. Is it differentiation, or something else, or what? Also how is it that multiplying the function by a number x (in this case) gives you the gradient at a point x? Shouldn't you have to differentiate the function to be able to work out the gradient at a certain point? Why are these functions different? What is going on here?
Thank you for your help.
 A: The notion of gradient involves calculus in an essential way. But we can make some progress without explicit use of the calculus. 
Consider the function $f(x)=a^x$, where $a\gt 1$. We want to find out how fast $f(x)$ is increasing at $x$. We can view this as the slope of the tangent line to the curve $y=a^x$ at the point $(x,a^x)$.  A sketch will show that $y=a^x$ is increasing very rapidly when $x$ is large.
One way of getting at the slope is to take a small positive number $h$, and calculate the slope of the line that goes through the two points $(x,a^x)$ and $(x+h,a^{x+h})$. For concreteness let $a=4$.
Our line therefore has slope 
$$\frac{4^{x+h}-4^x}{h}.$$
Note that this is equal to 
$$4^x \frac{4^h-1}{h}.$$
So the slope of the line is $4^x$ times a factor $\frac{4^h-1}{h}$ that does not depend on $x$.
The calculator can give useful information. Let $h$ be small, say $h=0.01$. Then $\frac{4^h-1}{h}\approx 1.39595$. So the gradient of $4^x$ is approximately $(4^x)(1.39595)$.  For a more accurate estimate of the gradient, take for example $h=0.001$. Then $\frac{4^h-1}{h}\approx 1.387256$. 
If we take even smaller values of $h$, our results can become somewhat unreliable, because of roundoff error. It turns out that the exact value of the gradient of $a^x$ is $(\ln a)a^x$.
For the level of accuracy ($2$ figures) in the table that you quote, and $a$'s of reasonable size, taking $h=0.01$ will give adequate estimates.   
A: Actually, none of the results given are correct.  The coefficients, 0.7, 1.1, 1.4, are rounded to two decimal places.
Given any a, the "difference quotient" for $y= a^x$ is $\frac{a^{x+ h}- a^x}{h}= \frac{a^xa^h- a^x}{h}= a^x\frac{a^h- 1}{h}$.  The "gradient" is the limit of that difference quotient as h goes to 0.  But we can factor out the $a^x$ and have $a^x\left(\lim_{h\to 0}\frac{a^h- 1}{h}\right)$.
The number "e" can be defined as the number such that that limit is 1: that is, $\lim_{h\to 0}\frac{a^h- 1}{h}= 1$. And ln(x) is the inverse function to $e^x$: $e^{ln(b)}= b$ for any positive b. Since $a^x= e^{ln(a^x)}= e^{xln(a)}$ it follows that the gradient of $a^x$ is $ln(a) a^x$.  In particular, if a= 2, ln(2)= 0.69314718055994530941723212145818 (approximately!) which is 0.7 to one decimal place, if a= 3, ln(3)= 1.0986122886681096913952452369225 which is 1.1 to one decimal place, and if a= 4, ln(4)= 1.3862943611198906188344642429164 which is 1.4 to one decimal place.
