When the master carpenter teaches his apprentice the trade, he always starts him off using a hand-saw and a brace-and-auger. After all, the kid's got to learn the basics. But after a while, the master has to show the apprentice how to use the power tools. What is to be gained cutting a sheet of plywood with a hand-saw anyway? Once you've done it with the electric saw, you'll never want to do it any other way.
Suppose you wanted to take the derivative of
f(x) = (x - 4)(x + 3)(x - 7)(x + 5) eq. 8.5-1
You could multiply out the polynomial and take the derivative
of the result, but that's a lot of work.
This is turning out to be a lot of work also. And it would be even worse if the original expression had more factors to it. The reason it's a lot of work is because the product rule, as useful as it is, is a hand tool. What we need is a power tool to attack this one.
Suppose you take the natural log of both sides of equation 8.5-1.
ln(f(x)) = ln((x - 4)(x + 3)(x - 7)(x + 5)) eq. 8.5-2aRemember the property we have for the product of logs:
ln(f(x)) = ln(x - 4) + ln(x + 3) + ln(x - 7) + ln(x + 5) eq. 8.5-2bNow apply the
f'(x) 1 1 1 1From here it's easy to solve for f'(x). Just multiply through by f(x). And we know what f(x) is because it was given in
= + + + eq. 8.5-3a f(x) x - 4 x + 3 x - 7 x + 5
æ 1 1 1 1 ö f'(x) = çEquation 8.5-3b is a perfectly good form for representing f'(x), but if you wanted to you could simplify it by multiplying through. Notice that each fraction cancels a different factor in the original f(x).
+ + + ÷ (x - 4)(x + 3)(x - 7)(x + 5) è x - 4 x + 3 x - 7 x + 5 ø eq. 8.5-3b
f'(x) = (x + 3)(x - 7)(x + 5) + (x - 4)(x - 7)(x + 5) + (x - 4)(x + 3)(x + 5) + (x - 4)(x + 3)(x - 7) eq. 8.5-3cThis procedure for taking the derivative of a multiple product is called logarithmic differentiation. The f(x) does not have to be the product of binomials to work. It can be the product of any kind of functions:
f(x) = x2e2xsin(x) eq. 8.5-4a ln(f(x)) = ln(x2) + ln(e2x) + ln(sin(x)) = 2ln(x) + 2x + ln(sin(x)) eq. 8.5-4b f'(x) 2 cos(x) 2Here are the steps for doing logarithmic differentiation of the product of multiple factors:
= + 2 + = + 2 + cot(x) eq. 8.5-4c f(x) x sin(x) x æ 2 ö f'(x) = ç + 2 + cot(x) ÷ x2e2xsin(x) eq. 8.5-4d è x ø f'(x) = 2x e2xsin(x) + 2x2e2xsin(x) + x2e2xcos(x) eq. 8.5-4e
After you've mastered the steps above, you can combine them all into one. Here's the plan. Suppose you have
f(x) = g1(x)g2(x)g3(x)g4(x) eq. 8.5-5That is, f(x) is, in this case, the product of four functions. There's nothing magical about four -- it could be the product of any number of functions. To find f'(x), simply write the product as many times as you have factors (in this case that's four times) with plus signs between them, but in each product take the derivative of just one of the factors. It has to be a different factor in each one that you take the derivative of. So here we would have:
f'(x) = g1'(x)g2(x)g3(x)g4(x) + g1(x)g2'(x)g3(x)g4(x) + g1(x)g2(x)g3'(x)g4(x) + g1(x)g2(x)g3(x)g4'(x) eq. 8.5-6For example, say
_ f(x) = (x2 - 3x + 7)ln(x)tan(x)Öx eq. 8.5-7Then following the abbreviated procedure, the derivative would be
_ 1 _ f'(x) = (2x - 3)ln(x)tan(x)Öx + (x2 - 3x + 7)
tan(x)Öx + x _ 1 (x2 - 3x + 7)ln(x)sec2(x)Öx + (x2 - 3x + 7)ln(x)tan(x) 2Öx eq. 8.5-8
1) Use logarithmic differentiation to find the derivative of
(x + a)(x - b) f(x) =where a, b, c, and d are all constants. Hint: Observe that this is the product of four functions. Identify them before you proceed.
(x + c)(x - d)
2) Use logarithmic differentiation (or use the shortcut method) to find the derivative of
x2e-xsin(2x) f(x) =Hint: Again observe that this is the product of four functions.
1 + x2
Note that Leibniz' Rule is not taught in many first year calculus classes. If that is the case in your class you can consider this to be optional material. You should, however, recognize that Leibniz' Rule is a very useful shortcut and it is not that difficult. So you might want to learn it anyway.
We go back to the derivative of the product of just two factors.
You learned a while back that to do that you use the
f(x) = x5sin(3x) f'(x) = 5x4sin(3x) + 3x5cos(3x) f"(x) = 20x3sin(3x) + 15x4cos(3x) + 15x4cos(3x) - 9x5sin(3x)Just the thought of doing the third derivative this way is already starting to make me feel tired. But notice that the second and third terms of the second derivative are the same. It's very much like when you take
(a + b)2 = a2 + ab + ab + b2and the second and third terms are repeated. Leibniz noticed the same thing back in the 17th century. Indeed he was able to show that the mechanics for taking the nth derivative of a product was entirely analogous to taking the nth power of a binomial. And that gives you a useful shortcut for taking high order derivatives of a product. Leibniz' rule is this: If
n ænö f(n)(x) = å èkø g(n-k)(x)h(n)(x) eq. 8.5-9 k=0where
ænö n! èkø =Or in other words, it's a binomial coefficient. Also g(n-k)(x) is the n-kth derivative of g(x). And h(n)(x) is the nth derivative of h(x). Note that the zeroth derivative of a function is the function itself.
Equation 8.5-9 might still look pretty opaque to you, so let's do an example. We'll take
the fourth derivative of
g(x) = x5 h(x) = sin(3x)
g'(x) = 5x4 h'(x) = 3 cos(3x)
g"(x) = 20x3 h"(x) = -9 sin(3x)
g(3)(x) = 60x2 h(3)(x) = -27 cos(3x)
g(4)(x) = 120x h(4)(x) = 81 sin(3x)
You also have the binomial coefficients for exponent of 4 are:
æ4ö æ4ö æ4ö æ4ö æ4ö
è0ø = 1 è1ø = 4 è2ø = 6 è3ø = 4 è4ø = 1
Applying the sum shown in equation 8.5-9, you have for the fourth derivative of
f(4)(x) = g(4)(x)h(x) + 4g(3)(x)h'(x) + 6g"(x)h"(x) + 4g'(x)h(3)(x) + g(x)h(4)(x)And substituting
f(4)(x) = 120x sin(3x) + 720x2 cos(3x) - 1080 x3 sin(3x) - 540 x4 cos(3x) + 81 sin(3x)
Here again is 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
where the bk in the nth row of the table is the binomial coefficient
ænö n! èkø =
Use Leibniz' Rule together with the table of binary coefficients above to find the first four derivatives of
e-x f(x) =
Step 1: This is the product of what two functions? Identify
them and call them g(x) and h(x).
Step 2: Find the first four derivatives of g(x) and h(x).
You will need these in order to apply Leibniz's Rule.
Step 3: Marry the products derivatives of g(x) and h(x)
to the appropriate coefficients in the table. The first derivative will use the
first row of the table, the second derivative the second row, and so on. Use the fourth
derivative that we did above as a guide.
This proof is optional material. You won't need to know it for class, and you don't even need to know it in order to be able to use Leibniz' Rule effectively. But if you want to know why Leibniz' rule works, here's why.
Leibniz' Rule is quite easy to prove by induction. You just have to be willing to pay close attention to indexed variables. You have to show that the coefficients that you end up with in front of each term have the same properties as the binomial coefficients. The properties that seal the proof is
ænö ænö è0ø = 1 ènø = 1and
æn+1ö ænö æ n ö è k ø = èkø + èk-1øIf cn,k is the coefficient for the kth term of the nth derivative of g(x)h(x), then we have to prove that
cn+1,0 = cn+1,n+1 = 1for any counting number, n, given that
cn,0 = cn,n = 1And we have to prove that
cn+1,k = cn,k + cn,k-1for any counting number, n, and for
With that in mind, consider that the nth derivative of g(x)h(x) must be something in the form of.
cn,0g(x)h(n)(x) + cn,1g'(x)h(n-1)(x) + cn,2g"(x)h(n-2)(x) + ... + cn,ng(n)h(x)At this point we don't have to know anything about what the values of the c's are.
To find the n+1st derivative, you take the derivative of the above.
To do that you must apply the
Here is the sum of all the left terms deriving from taking the derivative of the expression above:
cn,0g'(x)h(n)(x) + cn,1g"(x)h(n-1)(x) + cn,2g(3)(x)h(n-2)(x) + ... + cn,ng(n+1)h(x)And here is the sum of all the right terms deriving from taking the derivative of the expression above:
cn,0g(x)h(n+1)(x) + cn,1g'(x)h(n)(x) + cn,2g"(x)h(n-1)(x) + ... + cn,ng(n)h'(x)When you add both of these sums together you get the terms of the n+1st derivative of g(x)h(x). To add them, you gather like terms, and in this way you derive the cn+1,k coefficients from the cn,k coefficients. You will find that the term, cn,0g(x)h(n+1)(x) stands by itself, so
Likewise cn,ng(n+1)(x)h(x) stands by itself as well, so
With every other term, you find that a left term groups with a right term when you gather like terms. They always end up as
cn,k-1g(k)(x)h(n-k)(x) + cn,kg(k)(x)h(n-k)(x) left term right termSince these must add up to the kth term of the n+1st derivative, it is clear that you must have
cn+1,k = cn,k + cn,k-1which was what we had to prove to show that the c's behave in the same way as do binomial coefficients.
So we have proved that from the nth rung of the ladder we can get to the n+1st rung. You still have to prove that you can get to the first rung, but that's easy. We know that
æ1ö æ1ö è0ø = è1ø = 1Well you only have to prove that
c1,0 = c1,1 = 1And that is easy to do because it follows directly from the
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