The Derivative, Part 2

I ended my last post with the rather daunting definition of the derivative:

\[f'( x) \ =\lim _{h\rightarrow 0} \ \frac{f( x+h) -f( x)}{h}\]

I will now show how this definition can be used to find much simpler ways to calculate a derivative.

Let’s start with an example that we already know the answer to and is the simplest function we can think of, f(x) = c where c is some constant. You know that if the function does not change anywhere over the values of x, its rate of change (derivative) is zero. You see this if you plotted the function – it’s a horizontal line and a horizontal line has a gradient of zero. So f‘(x) = 0. Let’s see if the derivative definition gives us the same answer.

\[f'( x) \ =\lim _{h\rightarrow 0} \ \frac{f( x+h) -f( x)}{h} =\lim _{h\rightarrow 0} \ \frac{c-c}{h} \rightarrow \frac{0}{0}\]

Well that didn’t help much – we just got an indeterminate form 0/0. But as I said in my last post, there will always be some algebraic manipulation required to remove the problem.

One common method is to multiply the numerator and denominator by the same fraction. This does not change the value of the expression but if you use the right fraction, it removes the issue. In this case, multiply top and bottom by 1/h.

\[f'( x) \ =\lim _{h\rightarrow 0} \ \frac{\frac{1}{h}( c-c)}{\frac{1}{h}( h)} =\lim _{h\rightarrow 0} \ \frac{\frac{1}{h}( 0)}{1} =\lim _{h\rightarrow 0} \ \frac{0}{1} =0\]

Notice that by doing that, we got rid of h and we are left with 0/1 which is definitely 0. So the first rule of finding a derivative: if f(x) = c, then f‘(x) = 0.

Now let’s look at a more complex function, but again, one you know the answer to. The generic equation of a line is f(x) = mx + c where m and c are specific numbers: f(x) = 3x + 7 is an example. Again, from your studies of linear equations, you know this kind of function will plot as a straight line with a gradient of m. So we know that if f(x) = mx + c, then f‘(x) = m. Does our definition give the same result?

\[f'( x) \ =\lim _{h\rightarrow 0} \ \frac{m( x+h) +c-( mx+c)}{h} =\lim _{h\rightarrow 0} \ \frac{mx+mh+c-mx-c}{h} \ \] \[=\lim _{h\rightarrow 0} \ \frac{mh}{h} =\lim _{h\rightarrow 0} \ m=m\]

So when we enter in the particular function into the definition, then expand it, get rid of the terms that cancel, then cancel the common factor h, we again get rid of the dependency on h. We are left with the limit of a constant m as h approaches 0. But as m does not care what h does, the answer is just m – just what we expected. So we now know if f(x) = mx + c, then f‘(x) = m.

Next time, I will do the same thing but use functions for which we don’t know the answer.