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. 8.5-1You could multiply out the polynomial and take the derivative of the result, but that's a lot of work. Another alternative is: you could observe that this is the product of

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~~=~~~~+~~~~+~~~~+~~~~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~~+~~~~+~~~~+~~~~÷ (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

f(x) = xHere are the steps for doing logarithmic differentiation of the product of multiple factors:^{2}e^{2x}sin(x) eq. 8.5-4a ln(f(x)) = ln(x^{2}) + ln(e^{2x}) + ln(sin(x)) = 2ln(x) + 2x + ln(sin(x)) eq. 8.5-4b f'(x) 2 cos(x) 2~~=~~~~+ 2 +~~~~=~~~~+ 2 + cot(x) eq. 8.5-4c f(x) x sin(x) x æ 2 ö f'(x) = ç~~~~+ 2 + cot(x) ÷ x~~^{2}e^{2x}sin(x) eq. 8.5-4d è x ø f'(x) = 2x e^{2x}sin(x) + 2x^{2}e^{2x}sin(x) + x^{2}e^{2x}cos(x) eq. 8.5-4e

- Take the natural log of both sides, that is the log of
`f(x)`on the left and the log of the product on the right. - Apply the identity for the log of a product to the right -- that is the log of the product is the sum of the logs.
- Use the chain rule to take the derivative of both the
left and right sides. The right side is a sum, so you can take its derivative term by term.
The left side will always give you
`f'(x)/f(x)`. - Multiply both sides by
`f(x)`to isolate`f'(x)`on the left and an expression for`f'(x)`on the right. - Replace
`f(x)`with the original product that you started with. - Optionally multiply the original product by the sum to simplify.

After you've mastered the steps above, you can combine them all into one. Here's the plan. Suppose you have

f(x) = gThat is,_{1}(x)g_{2}(x)g_{3}(x)g_{4}(x) eq. 8.5-5

f'(x) = gFor example, say_{1}'(x)g_{2}(x)g_{3}(x)g_{4}(x) + g_{1}(x)g_{2}'(x)g_{3}(x)g_{4}(x) + g_{1}(x)g_{2}(x)g_{3}'(x)g_{4}(x) + g_{1}(x)g_{2}(x)g_{3}(x)g_{4}'(x) eq. 8.5-6

_ f(x) = (xThen following the abbreviated procedure, the derivative would be^{2}- 3x + 7)ln(x)tan(x)Öx eq. 8.5-7

_^{ }1 _ f'(x) = (2x - 3)ln(x)tan(x)Öx + (x^{2}- 3x + 7)~~tan(x)Öx +~~_{ }x^{ }^{ }_^{ }1 (x^{2}- 3x + 7)ln(x)sec^{2}(x)Öx + (x^{2}- 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~~(x + c)(x - d)~~

**2)** Use logarithmic differentiation (or use the shortcut method) to find the derivative of

x^{2}e^{-x}sin(2x) f(x) =~~1 + x~~^{2}

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
*first* derivative. What if you want to know the second,
third, or fourth derivatives and so on. You can end up applying the product
rule many times because each derivative will have more products in it than
the last. Take, for example,

f(x) = xJust 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^{5}sin(3x) f'(x) = 5x^{4}sin(3x) + 3x^{5}cos(3x) f"(x) = 20x^{3}sin(3x) + 15x^{4}cos(3x) + 15x^{4}cos(3x) - 9x^{5}sin(3x)

(a + b)and 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^{2}= a^{2}+ ab + ab + b^{2}

where^{ }n ænö f^{(n)}(x) = å èkø g^{(n-k)}(x)h^{(n)}(x) eq. 8.5-9 k=0

ænö n! èkø =Or in other words, it's a binomial coefficient. Also~~k!(n-k)!~~

Equation 8.5-9 might still look pretty opaque to you, so let's do an example. We'll take
the fourth derivative of ` x ^{5}sin(3x)`

g(x) = xYou also have the binomial coefficients for exponent of^{5}h(x) = sin(3x) g'(x) = 5x^{4}h'(x) = 3 cos(3x) g"(x) = 20x^{3}h"(x) = -9 sin(3x) g^{(3)}(x) = 60x^{2}h^{(3)}(x) = -27 cos(3x) g^{(4)}(x) = 120x h^{(4)}(x) = 81 sin(3x) Table 8.5-1

æ4ö æ4ö æ4ö æ4ö æ4ö è0ø = 1 è1ø = 4 è2ø = 6 è3ø = 4 è4ø = 1 Table 8.5-2Applying the sum shown in equation 8.5-9, you have for the fourth derivative of

fAnd substituting^{(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)

f^{(4)}(x) = 120x sin(3x) + 720x^{2}cos(3x) - 1080 x^{3}sin(3x) - 540 x^{4}cos(3x) + 81 sin(3x)

Here again is a table of binomial coefficients:

n: b |

where the `b _{k}` in the

ænö n! èkø =~~k!(n-k)!~~

Use Leibniz' Rule together with the table of binary coefficients above to find the first four derivatives of

e^{-x}f(x) =~~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

cfor any counting number,_{n+1,0}= c_{n+1,n+1}= 1

cAnd we have to prove that_{n,0}= c_{n,n}= 1

cfor any counting number,_{n+1,k}= c_{n,k}+ c_{n,k-1}

With that in mind, consider that the `n`th derivative of `g(x)h(x)` must
be something in the form of.

cAt this point we don't have to know anything about what the values of the_{n,0}g(x)h^{(n)}(x) + c_{n,1}g'(x)h^{(n-1)}(x) + c_{n,2}g"(x)h^{(n-2)}(x) + ... + c_{n,n}g^{(n)}h(x)

To find the `n+1`st derivative, you take the derivative of the above.
To do that you must apply the `c _{n,k}` terms are
constants). That will yield two terms for each term above. I shall call them the "left term"
and the "right term." A left term is what you get by taking one of the products and
taking the derivative of the left-hand factor of a term but leaving the right-hand factor alone.
Likewise the right term is what you get by taking the derivative of the right-hand factor
of a term but leaving the left-hand factor alone. The product rule tells us that when you
take the derivative of each term, it will yield a left term and a right term.

Here is the sum of all the left terms deriving from taking the derivative of the expression above:

cAnd here is the sum of all the right terms deriving from taking the derivative of the expression above:_{n,0}g'(x)h^{(n)}(x) + c_{n,1}g"(x)h^{(n-1)}(x) + c_{n,2}g^{(3)}(x)h^{(n-2)}(x) + ... + c_{n,n}g^{(n+1)}h(x)

cWhen you add both of these sums together you get the terms of the_{n,0}g(x)h^{(n+1)}(x) + c_{n,1}g'(x)h^{(n)}(x) + c_{n,2}g"(x)h^{(n-1)}(x) + ... + c_{n,n}g^{(n)}h'(x)

Likewise `c _{n,n}g^{(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

cSince these must add up to the_{n,k-1}g^{(k)}(x)h^{(n-k)}(x) + c_{n,k}g^{(k)}(x)h^{(n-k)}(x) left term right term

cwhich was what we had to prove to show that the_{n+1,k}= c_{n,k}+ c_{n,k-1}

So we have proved that from the `n`th rung of the ladder we can get
to the `n+1`st 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

cAnd that is easy to do because it follows directly from the_{1,0}= c_{1,1}= 1

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