# The Derivative, Part 3

Now that we have some confidence that the derivative definition gives correct results of functions that we know the answer to, let’s look at a functional form where the answer is not known.

Consider f(x) = x². As you know, this function plots as the standard parabola. The slope of a tangent line on this curve (its rate of change) is not constant, unlike the cases we have looked at before, but it depends on where we are on the curve:

$f'( x) =\lim _{h\rightarrow 0}\frac{f( x+h) -f( x)}{h} = \lim _{h\rightarrow 0}\frac{( x+h)^{2} -x^{2}}{h} =$ $\lim _{h\rightarrow 0}\frac{x^{2} +2xh+h^{2} -x^{2}}{h} =\lim _{h\rightarrow 0}\frac{h( 2x+h)}{h} =\lim _{h\rightarrow 0}( 2x+h) =2x$

So again, we do some algebraic manipulation that gets rid of the h in the denominator. Remember, as we are taking the limit as h approaches 0, the x is essentially treated as a constant. So the final answer is f‘(x) = 2x. Refering back to the graph, this satisfies the tangent line slopes at -1 and 1: f‘(-1) = -2, f‘(1) = 2. At any other point on the graph, just evaluate f‘(x) = 2x to find the rate of change of f(x) = x² at a particular x.

Now do you have to evaluate the definition for every different function you come across? Thankfully, the answer is no. Mathematicians have long ago done the hard work for you but because of the properties of limits, many general rules can be made. For example, if you know the derivative of a function, but what you have is the same function but multiplied by a constant, the derivative of this new function is just the same constant times the derivative of the old function. For example, we now know that for f(x) = x², f‘(x) = 2x. But what about g(x) = 3x²? Well, g‘(x) will just be 3 times the derivative of x², so g‘(x) = 6x.

So the rule is, if g(x) = af(x) where a is a constant number, then g‘(x) = af‘(x). Another generic rule is that the derivative of a sum of functions is the sum of the individual derivatives: If h(x) = f(x) + g(x), then h‘(x) = f‘(x) + g‘(x).

It turns out that if

$f( x) =ax^{n}$

where n is any real number except -1, then

$f'( x) =anx^{n-1}$

So to find the derivative in this case, you just multiply the function by n and reduce the value of the exponent by 1.

Next time, I will present a table of common derivatives and do some sample problems.