where we have used LIN with
Section 2.6 Using the Arithmetic of Derivatives – Examples
Subsection 2.6.1 Using the Arithmetic of Derivatives – Examples
In this section we illustrate the computation of derivatives using the arithmetic of derivatives — Theorems 2.4.2, 2.4.3 and 2.4.5. To make it clear which rules we are using during the examples we will note which theorem we are using:
We’ll start with a really easy example.
Example 2.6.1. .
Example 2.6.2. .
Continuing on from the previous example, we can use the product rule and the previous result to compute
where we have used the product rule PR with and
Example 2.6.3. .
In the same vein as the previous example, we can use the quotient rule to compute
where we have used the quotient rule QR with and
Now for a messier example.
Example 2.6.4. Some examples should be messy.
Differentiate
This problem looks nasty. But it isn’t so hard if we just build it up a bit at a time.
- First,
is the ratio of and we will be able to get the derivative of just by applying the quotient rule. The derivative, of is easy, so let’s work on - The function
is the linear combination and we will be able to get the derivative of just by applying linearity (Theorem 2.4.2). The derivative, of is easy. So let’s work of - The function
is the ratio we will be able to get the derivative of just by applying the special case the quotient rule (Corollary 2.4.6). The derivative of is easy. - So we have completed breaking down
into easy pieces. It is now just a matter of reversing the break down steps, putting everything back together, starting with the easy pieces and working up to Here goes. - We now have an answer. But we really should clean it up, not only to make it easier to read, but also because invariably such computations are just small steps inside much larger computations. Any future computations involving this expression will be a lot easier and less error prone if we clean it up now. Cancelling the
and the in gives
While the linearity theorem (Theorem 2.4.2) is stated for a linear combination of two functions, it is not difficult to extend it to linear combinations of three or more functions as the following example shows.
Example 2.6.5. Linearity of the derivative of three or more functions.
We’ll start by generalising linearity to three functions.
This gives us linearity for three terms, namely (just replacing upper case names by lower case names)
Just by repeating the above argument many times, we may generalise to linearity for terms, for any natural number
Similarly, while the product rule is stated for the product of two functions, it is not difficult to extend it to the product of three or more functions as the following example shows.
Example 2.6.6. Extending the product rule to more than two factors.
Once again, we’ll start by generalising the product rule to three factors.
This gives us a product rule for three factors, namely (just replacing upper case names by lower case names)
Observe that when we differentiate a product of three factors, the answer is a sum of three terms and in each term the derivative acts on exactly one of the original factors. Just by repeating the above argument many times, we may generalise the product rule to give the derivative of a product of factors, for any natural number
We can also write the above as
When we differentiate a product of factors, the answer is a sum of terms and in each term the derivative acts on exactly one of the original factors. In the first term, the derivative acts on the first of the original factors. In the second term, the derivative acts on the second of the original factors. And so on.
If we make then each of the terms on the right hand side of the above equation is the product of and exactly ’s, and so is exactly So we get the following useful result
This last result is quite useful, so let us write it as a lemma for future reference.
Lemma 2.6.7.
This immediately gives us another useful result.
Example 2.6.8. Derivative of .
We can now compute the derivative of for any natural number Start with Lemma 2.6.7 and substitute and
Again — this is a result we will come back to quite a few times in the future, so we should make sure we can refer to it easily. However, at present this statement only holds when is a positive integer. With a little more work we can extend this to compute where is any positive rational number and then any rational number at all (positive or negative). So let us hold off for a little longer. Instead we can make it a lemma, since it will be an ingredient in quite a few of the examples following below and in constructing the final corollary.
Lemma 2.6.9. Derivative of .
Back to more examples.
Example 2.6.10. Derivative of a polynomial.
Example 2.6.11. Derivative of product of polynomials.
In this example we’ll compute in two different ways. For the first, we’ll start with the product rule.
For the second, we expand the product first and then differentiate.
Example 2.6.12. Derivative of a rational function.
Example 2.6.13. Derivative of a cube-root.
In this example, we’ll use a little trickery to find the derivative of The trickery consists of observing that, by the definition of the cube root,
Since both sides of the expression are the same, they must have the same derivatives:
We already know by Theorem 2.2.4 that
Since we know that we must have
which we can rearrange to give the result we need
Example 2.6.14. Derivative of a positive rational power of .
In this example, we’ll use the same trickery as in Example 2.6.13 to find the derivative for any two natural numbers and By definition of the root,
That is, and are the same function, and so have the same derivative. So we differentiate both of them. We already know that, by Lemma 2.6.9 with
Remember that Now these two derivatives must be the same. So
and, rearranging things,
So finally
Notice that this has the same form as Lemma 2.6.9, above, except with allowed to be any positive rational number, not just a positive integer.
Example 2.6.15. Derivative of .
In this example we’ll use the quotient rule to find the derivative of for any natural number
Example 2.6.16. Derivative of a negative rational power of .
In this example we’ll use the quotient rule to find the derivative of for any pair of natural numbers and By the special case the quotient rule (Corollary 2.4.6) with and
Note that we have found, in Examples 2.2.2, 2.6.14 and 2.6.16, the derivative of for any rational number whether 0, positive, negative, integer or fractional. In all cases, the answer is
Corollary 2.6.17. Derivative of .
We shall show, in Example 2.10.5, that the formula in fact applies for all real numbers not just rational numbers.
Back in Example 2.2.9 we computed the derivative of from the definition of the derivative. The above corollary (correctly) gives
but with far less work.
Here’s an (optional) messy example.
Example 2.6.18. Optional messy example.
Find the derivative of
- As we seen before, the best strategy for dealing with nasty expressions is to break them up into easy pieces. We can think of
as the five–fold product - By now, the derivatives of the
’s should be easy to find: - Now, to get the derivative
we use the –fold product rule which was developed in Example 2.6.6, together with the special case of the quotient rule (Corollary 2.4.6). by is often useful in simplifying the derivative of a product of many factors.1
Also take a look at “logarithmic differentiation” in Section 2.10.
Exercises 2.6.2 Exercises
Exercises — Stage 1 .
Exercises — Stage 2 .
3.
Differentiate and factor the result.
4.
Differentiate
5.
Differentiate
6.
Differentiate
7. (✳).
Compute the derivative of
8.
9.
Differentiate
10. (✳).
Compute the derivative of
11. (✳).
Compute the derivative of
12. (✳).
Compute the derivative of
13. (✳).
14.
Differentiate
15.
Differentiate