Box 4.1 Coached Exercise: Finding a Derivative
© 1996 by Karl Hahn
The problem is to apply equation 4.1-2 in order to find
df f(x + h) - f(x) f'(x) =You should know what the function,
= lim eq. 4.1-2 dx h → 0 h
Step 1: Write expressions for
Step 2: Rewrite equation 4.1-2, but substitute your expressions for
Step 3: One of the expressions can be multiplied out into an expanded form. Rewrite the equation again using the expanded form.
Step 4: You should see a cancellation you can take in the numerator of the of your equation. Cancel it and rewrite the equation in the simplified form.
Step 5: Do you have a quotient of polynomials in h? If so, you can use the rule we discovered in section 2.5 to take the limit. Rememeber, find the lowest power of h whose numerator and denominator coefficients are not both zero. The ratio of those coefficients is your limit. And remember that a coefficient can be an expression involving x, but a coefficient cannot be an expression involving h. Why? Because we are dealing with polynomials in h. Furthermore, when you take the limit as h goes toward zero (or toward anything else, for that matter), you should always end up with an expression that contains no h's.
And if you don't have a quotient of polynomials in h then go back and check your work. You probably made an algebra mistake somewhere.
Does the expression you have derived for
If it does conform, then try one more check. Use your expression for
If you were unable to get through any step, go back to section 4.0 and review the method we used to find the exact grade of the animal's wall at any point. Then return here and try this exercise again.
Finding the Derivative of g(x) = xn
In order to find this derivative, we will be using the binomial
theorem. You probably covered it in algebra, but it is worth reviewing
once again. The binomial theorem gives you a formula for expanding the
(u + v)n = b0unv0 + b1un-1v1 + b2un-2v2 + ... + bn-1u1vn-1 + bnu0vn eq. 4b-1I know that thing in equation 4b-1 looks pretty nasty, but don't panic. It is simply the sum of
You will remember from algebra that anything to the zero power is 1, and anything to the first power is itself.
Do you remember how the counting numbers are like ladder? If you can get to the first rung, and if from any rung you can get to the next rung, then you can get to any rung. That is how we will approach the binomial problem.
The first rung is where
(u + v)1 = u + v = (1)u1v0 + (1)u0v1 eq. 4b-2This does seem to agree with equation 4b-1. It has
Now let's see what happens if we are poised on the nth rung
of the ladder and we try to climb to the
(u + v)n+1 = (u + v) (u + v)n eq. 4b-3On the right hand side of 4b-3 we can see that one of the factors is the same as the left hand side of 4b-1 as it applies to the nth rung of the ladder. Since we are assuming that 4b-1 already holds for that rung (because we are trying to show that 4b-1 holds for the
(u + v)n+1 = (u + v) (b0unv0 + b1un-1v1 + b2un-2v2 + ... + bn-1u1vn-1 + bnu0vn) eq. 4b-4
By multiplying through first by u, then by v, and then gathering up terms of like powers into columns, we get:
(u + v)n+1 = b0un+1v0 + b1unv1 + b2un-1v2 + ... + bn-1u2vn-1 + bnu1vn + b0unv1 + b1un-1v2 + ... + bn-2u2vn-1 + bn-1u1vn + bnu0vn+1 eq. 4b-5All the terms that derived from multiplying by u are in the first line after the equal sign; all the terms that derived from multiplying by v are in the second line after the equal sign. They are lined up according to like powers.
First, notice that we are definitely going to get different coefficients
b*0 = b0 eq. 4b-6a b*n+1 = bn eq. 4b-6band for all the subscripts in between:
b*k = bk-1 + bk eq. 4b-6c
What we have shown so far is that 4b-1 holds for all counting numbers,
n, except with a different set of coefficients for each n.
We also know exactly what the coefficients are for
Using that knowledge, we can develop a table of binomial coefficients:
n: b0 b1 b2 b3 b4 b5 b6 b7 b8 b9 b10 b11 b12 1: 1 1 2: 1 2 1 3: 1 3 3 1 4: 1 4 6 4 1 5: 1 5 10 10 5 1 6: 1 6 15 20 15 6 1 7: 1 7 21 35 35 21 7 1 8: 1 8 28 56 70 56 28 8 1 9: 1 9 36 84 126 126 84 36 9 1 10: 1 10 45 120 210 252 210 120 45 10 1 11: 1 11 55 165 330 462 462 330 165 55 11 1 12: 1 12 66 220 495 792 924 792 495 220 66 12 1
You can see for yourself how table how the table was formed. We know the
first line, when
There is more to know about binomial coefficients, and you can (and should)
learn it by
But for the purposes of finding the derivative of
So, with our knowledge of binomial coefficients in place, let's now proceed
to finding the deriviative of
Step 1: Find expressions for
Step 2: Substitute those expressions into 4.1-2 in place of
(x + h)n - xn g'(x) = lim
eq. 4b-7 h → 0 h
Step 3: Apply the binomial theorem to
1xnh0 + nxn-1h1 + b2xn-2h2 + ... + bn-1x1hn-1 + bnx0hn - xn g'(x) = lim
h → 0 h eq. 4b-8
Step 4: Take the cancellation. Notice that
1xnh0 is the same as the xn
at the end of the numerator, except that it is multiplied by some things
that are each equal to 1. Hence,
nxn-1h1 + b2xn-2h2 + ... + bn-1x1hn-1 + bnx0hn g'(x) = lim
eq. 4b-9 h → 0 h
Step 5: Use the rule to take the limit. The rule we discussed in section 2.5 applies here because we have a polynomial in h in the numerator and a polynomial in h in the denominator. The polynomial in the numerator has 1 as its lowest power of h and the coefficient of that term is nxn-1. The polynomial in the denominator has only one term, and it has a coefficient of 1 and a power of 1 (remember that h1 is the same as h). At least one of the coefficients is not zero. So the rule says that the limit is the quotient of these.
g'(x) = nxn-1 eq. 4b-10According to the rule from section 2.5, all the higher power terms of h (that is with exponents of 2 or greater) with all their bk coefficients (that we didn't bother to attach values to) make not a bit of difference when you take the limit as h goes toward zero. In the limit, they all drop away. Why? Because the lowest power in the denominator is 1. All the numerator terms with greater power than that don't do squat in the limit.
The fact that the derivative of xn is always nxn-1 is worth memorizing because it comes up often. But don't go forgetting how it is derived. And remember too that this derivation applies only to the case where n is a counting number.
And what, you might ask, about the case when n is not a counting
number. What if it is a fraction, such as 1/2 (note that
Use the formula given in 4b-10 to find the derivatives of the following:
f(x) = x3 f(x) = x4 f(x) = x5When you're done, try plugging each of these functions into the limit formula given in 4.1-2. Expand each using the binomial theorem (you can copy the binomial coefficients you need from table 4b-1 if you like), then take the limit using the rule from section 2.5. This set of derivatives ought to agree with the first (i.e. the ones you got by applying 4b-10).
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