Now with our knowledge of limits in place, we are finally in a position to do some calculus. As mentioned earlier, the starting point of calculus is the notion of a derivative. So what is a derivative? The derivative is a property of a function. Every sufficiently smooth function has a derivative. Intuitively, the derivative of a function f(x) is a new function, denoted as f'(x), whose value f(x0) at every value of x0 in its domain is equal to the slope of f(x) at x = x0. So what do we mean by the slope of f(x) at x = x0? This is defined to be the slope of the tangent line to the curve y = f(x) at the point P0 = (x0, f(x0)). How do we find the tangent line? Let h be a (small) positive number, let x1 = x0 + h, and let P1 = (x1, f(x1)). Now consider the line passing through the points P0 and P1. What is the slope of this line? From Section 1.2, we know how to compute the slope of a line passing through two known points; the slope in this case is equal to
m = [f(x1) - f(x0)] / (x1 - x0) = [f(x0 + h) - f(x0)] / h.
Now we define the derivative of f at x0 to be the limit of m as h goes to zero, since this is equal to the slope of the tangent line. Specifically we have
Now since there is nothing special about the point x0, we can replace it by x, obtaining the following general definition of the derivative:
The line connecting the points P0 and P1 is known as a secant line. The following diagram show how the secant lines approach the tangent line at (x, f(x)) as h approaches zero. The secant lines are shown in purple and the tangent line in brown.
Figure 11.2.2: Secant Lines approaching the Tangent Line
(Image taken from Wikipedia)
We must be careful with Equation (11.2.1), for the derivative of a function is not always defined. For a given function f(x), the derivative f'(x0) only exists at x0 if the two-sided limit in (11.2.1) exists at x = x0. In this case, f is said to be differentiable at x0. If f(x) is not continuous at x0, then it is not differentiable there either. But f(x) may be continuous but not differentiable at x0. A good example of this is the absolute value function, which is continuous everywhere but not differentiable at x=0. The point x=0 is known as a cusp of the absolute value function. There are even functions which are continuous everywhere but differentiable nowhere!
Several different types of notation are used for the derivative. The derivative f'(x) of a function f(x) is also written as Df(x) and df / dx.
Theorem 11.2.3: The derivative satisfies the following properties. (Below, f and g are both functions and k is a constant.)
- (a) D(f + g) = Df + Dg
- (b) D(kf) = k Df
In other words, the derivative D is what is known as a linear operator.
- (a) Let F = f + g. We must show that F'(x) = f'(x) + g'(x). From (11.4.1) we have
- (b) Let F = kf. We must show that F'(x) = k f(x). We have
Theorem 11.2.4: Let f1(x), f2(x),..., fn(x) be a collection of n functions, let k1, k2,... ,kn be n constants, and let F(x) = k1f1(x) + k2f2(x) + ... + knfn(x). Then we have F'(x) = k1f1'(x) + k2f2'(x) + ... + knfn'(x).
Proof: We will prove this theorem by induction. First consider the base case, n=1. This case reduces to part (b) of Theorem 11.2.3. Next, we perform the induction step. We assume Theorem 11.4.3 holds for a particular positive integer n and prove that it holds for n+1. Here we have F(x) = k1f1(x) + k2f2(x) + ... + knfn(x) + kn+1fn+1(x). Let f(x) = k1f1(x) + k2f2(x) + ... + knfn(x) and let g(x) = kn+1fn+1(x). By part (a) of Theorem 11.2.3, we have F'(x) = f'(x) + g'(x). Now by our induction hypothesis, we have f'(x) = k1f1'(x) + k2f2'(x) + ... + knfn'(x) and by part (b) of Theorem 11.2.3 we have g'(x) = kn+1fn+1'(x). Now by part (a) of Theorem 11.2.3, we have F'(x) = f'(x) + g'(x), which we see is equal to k1f1'(x) + k2f2'(x) + ... + knfn'(x) + kn+1fn+1'(x). This proves the induction step and completes our proof.
In the remainder of this chapter, we will derive formulas for the derivatives of some simple functions, including polynomials and rational functions.
Theorem 11.2.5 (Power Rule): Let n be a nonnegative integer. The derivative of f(x) = xn is given by f'(x) = nxn-1. In other words, we have
- (11.2.6) d(xn)/dx = nxn-1.
Proof: To prove (11.2.6), we use the Binomial Theorem. We have
From Theorems 11.2.4 and 11.2.5, we can now derive a formula for the derivative of a univariate polynomial.
Theorem 11.2.7: Let p(x) = anxn + an-1xn-1 + ... + a2x2 + a1x + a0 be an arbitrary polynomial of degree n, for some nonnegative integer n. Then we have p'(x) = nanxn-1 + (n-1)an-1xn-2 + ... + 2a2x+ a1.
Proof: This result follows immediately from Theorems 11.2.4 and 11.2.5 by letting fm(x) = xm and km = am for 0≤m≤n.
Example 1: Compute the derivatives of the following polynomials:
- (a) f(x) = x2 + 3x + 5
- (b) f(x) = 3x - 4
- (c) f(x) = x10 - 4x6 + 7x5 - 12x3 + 22x - 30
- (a) Using Theorem 11.4.5, we see that f'(x) = 2x + 3.
- (b) Using Theorem 11.4.5, we see that f'(x) = 3. (Note that linear polynomials have constant derivatives since they have constant slope.)
- (c) Using Theorem 11.4.5, we see that f'(x) = 10x9 - 24x5 + 35x4 - 36x2 + 22.
Like any other functions, derivatives may be evaluated for any particular value of their argument. This tells us the slope of the function in question at particular values of x, which can be quite useful information.
Example 2: Compute the slopes of each of the polynomials from Example 1 at x = 0 and at x = 10.
- (a) We have f'(0) = (2)(0) + 3 = 3 and f'(10) = (2)(10) + 3 = 23.
- (b) We have f'(0) = f'(10) = 3.
- (c) We have f'(0) = 22 (constant term of f'(x)) and f'(10) = 10*109 - 24*105 + 35*104 - 36*102 + 22 = 9,997,946,422.