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.
Best Answer
You seem to have the right idea.
$$1\text{ e+11}=1 \times 10^{+11}$$
You should know that $10$ raised to any positive integer is a $1$ with that many $0$s behind it. So
$$10^{+11}=100000000000$$