INVESTIGATION - Differentiation and exponential functions

Lesson

Derivative of an exponential function

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

f'\left(x\right)	=	\lim_{h\rightarrow 0}\frac{f\left(x+h\right)-f\left(x\right)}{h}
	=	\lim_{h\rightarrow 0}\frac{a^{x+h}-a^{x}}{h}
	=	\lim_{h\rightarrow 0}\frac{a^{x}{\left (a^{h}-1 \right )}}{h}	The term a^x does not depend on h and can be brought to the from of the expression
	=	a^{x}\times \lim_{h\rightarrow 0}\frac{a^h-1}{h}

Thus, provided the quantity \frac{a^h-1}{h} converges as h\rightarrow 0, that is the limit exists, the derivative of an exponential function exists and is the product of the original function and some constant whose value depends on the base a.

Activity

Let's investigate if this limit exists for different values of a. A convenient way to investigate this will be to set up a spreadsheet.

Step 1. In the first column make a list of values for h that approach zero.

Step 2. Decide on some values of a through which to explore the limit. These should be positive numbers.

Step 3. In cell B2 enter the formula to calculate \frac{a^h-1}{h}, where a is the value in the first row and h is the value in the first column. So for this cell we would have the formula =(B1^A2-1)/A2.

Step 4. So we can drag the formula, rather than rewrite it for each cell, we can lock in a column or row by placing the symbol \$ before a letter or number for a cell code to lock in the column or row respectively. Alter the formula in cell B2 to =(B$1^$A2-1)/$A2 and then drag the formula to the other cells in the table.

Discussion questions

Do each of the values appear to approach a limit?
Do they approach the same limit if h approaches zero from below? Change the h values to negatives to check this.
Try different values for a including fractions and irrational numbers. Do they all appear to approach a limit?

Applying to derivatives

In fact, for the function f\left(x\right)=2^x, the limit \lim_{h\rightarrow 0}\frac{2^h-1}{h} does exist and converges to an irrational number that is approximately 0.693147. This means the function f\left(x\right)=2^x, has the following derivative: f'\left(x\right)\approx 0.693147\times 2^x, since we recall above that f'\left(x\right)=a^{x}\times \lim_{h\rightarrow 0}\frac{a^h-1}{h}.

The table shows derivative results for four values of a:

a	\lim_{h\rightarrow 0}\frac{a^h-1}{h}	f'(x)
2	0.693147...	\approx 0.693147\times 2^x
3	1.098612...	\approx 1.098612\times 3^x
4	1.386294...	\approx 1.386294\times 4^x
5	1.609438...	\approx 1.609438\times 5^x

Do the values in the table match your approximation to the limit you found using a spreadsheet?

The results suggest that there is a base between a=2 and a=3 such that the quantity in the middle column - namely \lim_{h\rightarrow 0}\frac{a^h-1}{h}, is exactly 1.

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

Our first investigation centred around the number e, which happens to lie in between 2 and 3, and hinted at special properties of functions base e in calculus. So it may not come as a great surprise that the function in question is f\left(x\right)=e^x.

Thus with the base e\approx 2.718, we have the important result:

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

Check

Using your spreadsheet confirm that for e the \lim_{h\rightarrow 0}\frac{a^h-1}{h} appears to be equal to 1. We can prove this is in fact the case once we have looked more closely at logarithms.

Another view

We have seen that the derivative of f\left(x\right)=a^x is given by f'\left(x\right)=a^{x}\times \lim_{h\rightarrow 0}\frac{a^h-1}{h}. If we look closely at the limit in this derivative \lim_{h\rightarrow 0}\frac{a^h-1}{h} we can interpret this as the gradient of the tangent to the function at x=0, since:

f'\left(x\right)	=	\lim_{h\rightarrow 0}\frac{f\left(x+h\right)-f\left(x\right)}{h}
\therefore f'\left(0\right)	=	\lim_{h\rightarrow 0}\frac{f\left(0+h\right)-f\left(0\right)}{h}
	=	\lim_{h\rightarrow 0}\frac{a^{\left(0+h\right)}-a^0}{h}
	=	\lim_{h\rightarrow 0}\frac{a^h-1}{h}

Hence, the derivative of f\left(x\right)=a^x is f'\left(x\right)=f'\left(0\right)\times a^x. And here we could experiment to find a value for a such that the tangent to the function at x=1 has a gradient of 1. Then again we would have discovered a function that is its own derivative.

Experiment with the applet below by adjusting the value of a to find the value that has this property to 2 decimal places. Again, while this doesn't prove that e is the number in question, it does offer further confirmation that it is a good candidate and we can prove this fact once we have further tools.

Created with Geogebra

Comparison

The following applet shows the function f\left(x\right)=a^x and its derivative for a range of values between a=2 and a=3 . Satisfy yourself that the curves of the function and the differential function coincide when a\approx 2.72. That is f\left(x\right)=f'\left(x\right) for approximately this value.

Created with Geogebra

What does this all mean?

What does a function being its own derivative translate to? As well as simplifying many calculations in calculus this gives the graph the following properties:

This means the value of the function at any point on the graph is equal to the gradient of the tangent at that point
Equivalently the value of the function and the instantaneous rate of change are the same for any given value of x
We also have the value of the function at any point is equal to the area under the curve up to that point