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# Explore properties of an alternative definition of the derivative

Consider the following different definition of the a derivative:

where means .

1. Find formulas for the “derivative” of the sum, difference, product, and quotient of functions.
2. Find an expression for in terms of the usual derivative .
3. Find the functions for which .

1. Sum. Using the formula in the alternative definition,

Now, for the remaining limit term we multiply by (thanks to abcd in the comments for the suggestion!). So, we then have

Now, for the remaining limit, we use the product rule for the derivative in this alternative definition, which we derive below. (This last limit is . Below we derive that .) Therefore, we have

Difference. This follows just as the above derivation for except that we get minus signs:

Product. Using the alternative definition of the “derivative” we compute,

2. Expressing in terms of the usual derivative we have,

We used the same idea we used in deriving some of the above formulas that , and that is a constant in , so we can pull it out of the limit.

3. Now, to find the function such that we set the two expressions equal and solve for (using from part (c)),

In either case we have for some constant .

1. Hteica says:

Hi, in (a), I don’t think we are allowed to multiply [ (fg)(x+h) + (fg)(x) ] / [ (fg)(x+h) + (fg)(x) ], what happens when f is defined to be -1 for all values except at x where f(x)=1, and g defined as constant 1 for any values (but the limit of D*f and D*g do exist), we get (fg)(x+h) + (fg)(x)= -1+1=0, which means we were multiplying by 0/0 all along, which is not ok. Thanks!

2. abcd says:

Also, I believe that your solution for part a (formula for ) is wrong. In your solution, you reduced to two different expressions: , and .

If both are true, we can get a different expression for , e.g. which is clearly different from the answer.

I believe the error was made when you cancelled and similar terms in the fraction because for the limit it is illegal to cancel a term in only A (or only B) if B (or A) diverges.

• RoRi says:

Hey, thanks for this and the other comment. You’re right, my solution contains the error you mention. Fixing it now.

3. abcd says:

For part a, I think that we can deal with the term by multiplying by to give

which reduces to .

Applying the product rule derived later reduces the quotient, thus skipping the painful manipulation.

• Awamoki says:

abcd, there is another way to prove the part a without applying the product rule. We can add and subtract (f(x)^2)(g(x+h)^2) at the numerator after you multiplied by [ (fg)(x+h) + (fg)(x) ] / [ (fg)(x+h) + (fg)(x) ] and rearrange the sum.

4. tasos says:

Alternatively for part a, is continuous and from composition is differentiable and continuous hence we can get from part b (where apostol implies differentiability of ) to make the calculations much more painless. The direct results are recursive formulas of normal derivatives rather than though.

• RoRi says:

Yes, that works also and is decidedly less painful. I figured since Apostol put it before part (b) that he wanted us to suffer and compute it out the long way by hand to see how it works. Maybe I overestimate how mean he is though.

PS. I hope you don’t mind that I edit LaTeX into your comments. Let me know if it’s a problem or if I introduce any errors.

• tasos says:

I’m sure that’s what he had in mind. And no, I honestly appreciate you introducing latex into my comments.

• Mihajlo says:

It doesn’t seem to be the matter of meanness, but that you can learn more if you do a) without using the result of b).