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.

Gradient table

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.

  • $\begingroup$ @AndréNicolas, so the construction of this table requires knowledge of calculus? Is this the only way it can be derived? $\endgroup$ – seeker Nov 9 '13 at 21:34
  • $\begingroup$ I see, but how is it in this case they got $\frac{{{4^h} - 1}}{h}$ to equal $1.4$, is there a certain value of h that should be recommended? $\endgroup$ – seeker Nov 9 '13 at 21:52
  • $\begingroup$ I've got it now, thank you so much! It's a bit hard to award you the answer, so if you want, just compose a little something so I can award it. $\endgroup$ – seeker Nov 9 '13 at 22:00
  • $\begingroup$ It's OK, you are likely to get some answers soon. I will delete comments so as not to discourage answers, maybe you can do the same. I will be away, will write an answer in a few hours in the unlikely case none is given. $\endgroup$ – André Nicolas Nov 9 '13 at 22:03

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.


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.


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