The Power Rule

Now let us see how, on first principles, we can differentiate some simple algebraical expression .

Case of a Positive Power

Example 1.

Example 4.1. Differentiate .

Solution. Let us begin with the simple expression . Now remember that the fundamental notion about the calculus is the idea of growing. Mathematicians call it varying. Now as  and  are equal to one another, it is clear that if  grows,  will also grow. And if  grows, then  will also grow. What we have got to find out is the proportion between the growing of  and the growing of . In other words, our task is to find out the ratio between  and , or, in brief, to find the value of .

Let , then, grow a little bit bigger and become ; similarly,  will grow a bit bigger and will become . Then, clearly, it will still be true that the enlarged  will be equal to the square of the enlarged . Writing this down, we have: Doing the squaring we get:

What does mean? Remember that meant a bit—a little bit—of . Then  will mean a little bit of a little bit of ; that is, as explained above, it is a small quantity of the second order of smallness. It may therefore be discarded as quite inconsiderable in comparison with the other terms. Leaving it out, we then have: Now ; so let us subtract this from the equation and we have left Dividing across by , we find

Now this1 is what we set out to find. The ratio of the growing of to the growing of is, in the case before us, found to be .

Suppose and therefore . Then let grow till it becomes (that is, let ). Then the enlarged  will be . But if we agree that we may ignore small quantities of the second order,  may be rejected as compared with ; so we may round off the enlarged  to .  has grown from to ; the bit added on is , which is therefore .

. According to the algebra-working of the previous paragraph, we find . And so it is; for and .

But, you will say, we neglected a whole unit.

Well, try again, making a still smaller bit.

Try . Then , and

Now the last figure is only one-millionth part of the , and is utterly negligible; so we may take without the little decimal at the end. And this makes ; and , which is still the same as .

Example 2.

Example 4.2. Try differentiating in the same way.

Solution. We let grow to , while  grows to .

Then we have

Doing the cubing we obtain

Now we know that we may neglect small quantities of the second and third orders; since, when  and  are both made indefinitely small,  and  will become indefinitely smaller by comparison. So, regarding them as negligible, we have left:

But ; and, subtracting this, we have: and

Example 3.

Example 4.3. Try differentiating .

Solution. Starting as before by letting both  and  grow a bit, we have: Working out the raising to the fourth power, we get Then striking out the terms containing all the higher powers of , as being negligible by comparison, we have Subtracting the original , we have left and

Now all these cases are quite easy. Let us collect the results to see if we can infer any general rule. Put them in two columns, the values of  in one and the corresponding values found for  in the other: thus

Just look at these results: the operation of differentiating appears to have had the effect of diminishing the power of  by  (for example in the last case reducing to ), and at the same time multiplying by a number (the same number in fact which originally appeared as the power). Now, when you have once seen this, you might easily conjecture how the others will run. You would expect that differentiating would give , or differentiating would give . If you hesitate, try one of these, and see whether the conjecture comes right.

Example 4.

Example 4.4. Try differentiating .

Solution. Then Neglecting all the terms containing small quantities of the higher orders, we have left and subtracting leaves us whence exactly as we supposed.

Following out logically, our observation, we should conclude that if we want to deal with any higher power—call it —we could tackle it in the same way.

Let then, we should expect to find that For example, let , then ; and differentiating it would give .

And, indeed, the rule that differentiating gives as the result is true for all cases where is a whole number and positive. [Expanding by the binomial theorem (see the appendix) will at once show this.] But the question whether it is true for cases where has negative or fractional values requires further consideration.

Case of a Negative Power

Example 5.

Example 4.5. Differentiate .

Solution. We can write . Then proceed as before: Expanding this by the binomial theorem (see the appendix), we get So, neglecting the small quantities of higher orders of smallness, we have: Subtracting the original , we find or And this is still in accordance with the rule inferred above.

Case of a Fractional Power

Example 6.

Example 4.6. Differentiate .

Solution. Notice that .2 So let . Then, as before, Subtracting the original , and neglecting higher powers we have left: and . Agreeing with the general rule.

Summary

Let us see how far we have got. We have arrived at the following rule: To differentiate , multiply by the power and reduce the power by one, so giving us  as the result.

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Exercises

Differentiate the following:

Exercise 1.

Exercise 4.1.

 

Solution

 

 

 

Exercise 2.

Exercise 4.2.

 

Solution

 

 

 

Exercise 3.

Exercise 4.3.

 

Solution

 

 

Exercise 4.

Exercise 4.4.

 

Solution

 

 

 

Exercise 5.

Exercise 4.5.

 

Answer

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Solution

 

 

Exercise 6.

Exercise 4.6.

 

Answer

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Solution

 

 

 

Exercise 7.

Exercise 4.7.

 

Answer

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Solution

 

 

 

Exercise 8.

Exercise 4.8.

 

Solution

 

 

 

Exercise 9.

Exercise 4.9.

 

Answer

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Solution

 

 

 

Exercise 10.

Exercise 4.10.

 

Answer

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Solution

 

 

 


This ratio is the result of differentiating with respect to . Differentiating means finding the derivative. Suppose we had some other function of , as, for example, . Then if we were told to differentiate this with respect to , we should have to find , or, what is the same thing, . On the other hand, we may have a case in which time was the independent variable (see here), such as this: . Then, if we were told to differentiate it, that means we must find its derivative with respect to . So that then our business would be to try to find , that is, to find .↩︎

In general .↩︎