NZ Level 8 (NZC) Level 3 (NCEA) [In development]
Limit of (a^h-1)/h as h→0 (investigation)
Lesson

## First principles derivation

The derivative of the exponential function given by $f\left(x\right)=a^x$f(x)=ax can be developed by first principles:

 $f'\left(x\right)$f′(x) $=$= $\lim_{h\rightarrow0}\frac{f(x+h)-f(x)}{h}$limh→0f(x+h)−f(x)h​ $=$= $\lim_{h\rightarrow0}\frac{a^{x+h}-a^x}{h}$limh→0ax+h−axh​ $=$= $\lim_{h\rightarrow0}\frac{a^x\left(a^h-1\right)}{h}$limh→0ax(ah−1)h​ $=$= $a^x\times\lim_{h\rightarrow0}\frac{a^h-1}{h}$ax×limh→0ah−1h​

Thus, provided the quantity $\frac{a^h-1}{h}$ah1h converges as $h\rightarrow0$h0, the derivative exists as the product of the original function itself and some constant whose value depends on the base $a$a.

In fact, for the function $f\left(x\right)=2^x$f(x)=2x, as $h\rightarrow0$h0, the quantity $\frac{2^h-1}{h}$2h1h does converge to an irrational number that is approximately $0.693147$0.693147. This means that  $f\left(x\right)=2^x$f(x)=2x, then $f'\left(x\right)\approx0.693147\times2^x$f(x)0.693147×2x

The table shows derivative results for four values of a:

$a$a $\lim_{h\rightarrow0}\frac{a^h-1}{h}$limh0ah1h $f'(x)$f(x)
$2$2 $0.693147...$0.693147... $\approx0.693147\times2^x$0.693147×2x
$3$3 $1.098612...$1.098612... $\approx1.098612\times3^x$1.098612×3x
$4$4 $1.386294...$1.386294... $\approx1.386294\times4^x$1.386294×4x
$5$5 $1.609438$1.609438 $\approx1.609438\times5^x$1.609438×5x

The results suggest that there is a base between $a=2$a=2 and $a=3$a=3 where the quantity in the middle column - namely $\lim_{h\rightarrow0}\frac{a^h-1}{h}$limh0ah1h, is exactly $1$1.

This would reveal the existence of a function that is its own derivative!

Your turn now.  We can see that this function must exist somewhere between 2 and 3.  Pick three values between 2 and 3 and see if you can get closer to this function.

### In fact...

This remarkable and irrational base has the approximate value of $2.718281828459045$2.718281828459045 and has been denoted by the letter $e$e (standing for the word exponential, and first discovered by the mathematician Leonard Euler)

Thus with the base $e\approx2.718$e2.718, we have the important result:

If $f\left(x\right)=e^x,$f(x)=ex, then $f'\left(x\right)=e^x$f(x)=ex.

### Applet

The following applet shows the function $f\left(x\right)=a^x$f(x)=ax and its derivative for a range of values between $a=2$a=2 and $a=3$a=3 . Satisfy yourself that the curves of the function and the differential function coincide when $a\approx2.72$a2.72

### Outcomes

#### M8-11

Choose and apply a variety of differentiation, integration, and antidifferentiation techniques to functions and relations, using both analytical and numerical methods

#### 91578

Apply differentiation methods in solving problems