Is There a Simpler Way to Calculate Derivatives?

[cam875]

I have been studying calculus for only a few days now since I picked up a book on it, trying to get ahead for next year and it's really starting to sink in now and I decided to write down a different way for calculating the derivative, it isn't revolutionary, i am just wondering if it will be accepted by most people or if they prefer the traditional way. Its exactly the same as the traditional difference quotient but I switched it around because it made more sense this way to me.

$$f\prime(x)=\[ \lim_{x_{1} \to x_{2}} \frac{f(x_{2})-f(x_{1})}{x_{2}-x_{1}}$$

 

[mathman]

Since it is mathematically exactly equal to the usual approach, use it if it makes you happy.

 

[de_brook]

I have been studying calculus for only a few days now since I picked up a book on it, trying to get ahead for next year and it's really starting to sink in now and I decided to write down a different way for calculating the derivative, it isn't revolutionary, i am just wondering if it will be accepted by most people or if they prefer the traditional way. Its exactly the same as the traditional difference quotient but I switched it around because it made more sense this way to me.

$$f\prime(x)=\[ \lim_{a \to x} \frac{f(x)-f(a)}{x-a}$$

normally, a is a constant and x is a variable thus x tends to a not a tending to x, so u can reverse $$\{a \to x}$$

 

[cam875]

what if I made "a" a different variable could that fix it?

 

[Feldoh]

what if I made "a" a different variable could that fix it?

The way you defined it f is a function of x, not a it doesn't matter what you use you can't have a constant approaching a variable because for obvious reasons it just doesn't make sense.

 

[HallsofIvy]

As mathman told you originally, it doesn't matter. You are simply swapping letters.

Now, if you had a problem like "what is the derivative of f(x)= x² at x= a", "x" and "a" are already given meanings so you cannot be so cavalier about swapping them.

That is, the standard definition would give 2a while your formula would give 2x. They are exactly the same if you remember that you have swapped a and x. In other words, to answer the problem as given you would have to swap back to get 2a.

 

[cam875]

ok well I meant for "a" to be a variable, didn't realize it was a constant. Anyways I got it cleared up now, thanks guys.