INVESTIGATION - Differentiation and exponential functions

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 front 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

Follow the steps to investigate whether this limit exists for different values of a. A convenient way to do this is to set up a spreadsheet.

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

Step 2: Choose specific values of a  to investigate its 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 \frac{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 \frac{\$B1^{\$A2}-1}{\$A2}  and then drag the formula to the other cells in the table.

 

Discussion questions

1. Do each of the values appear to approach a limit?
2. Do they approach the same limit if h approaches zero from below? Change the h values to negatives to check this.
3. 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 this derivative: f'\left(x\right)\approx 0.693147\times 2^x, since  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.

 

The first investigation centred around the number e, which happens to lie in between 2 and 3, and hinted at special properties of functions based on 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 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}

 

Therefore, 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 determine a value for a such that the tangent to the function at x=0 has a gradient of 1. Then again, we would have discovered a function that is its own derivative.

Experiment with the applet provided by adjusting the value of a to determine the value that has this property rounded to two decimal places. Again, while this does not 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 next applet shows the function f\left(x\right)=a^x and its derivative for a range of values between a=2 and a=3 . Verify that the curves of the function and its derivative 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 these 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.

Examples

The value of the function at x=0 is 1. The gradient of the tangent at this point is 1.	The area under the curve up to this point is 1.

The value of the function at x=1 is e. The gradient of the tangent at this point is e.	The area under the curve up to this point is e.

The value of the function at x\approx 1.609 is 5. The gradient of the tangent at this point is 5.	The area under the curve up to this point is 5.

 

Using graphing software, you can investigate these properties further.