Connecting the Keystone Equation and Derivatives

[jedson303]

Here's a question. This formula seems to be the keystone of calculus.

View attachment 644That seems to be what the calculus books say, and it makes sense to me, as a rank beginner. This equation is what makes the seeming magic of defining the slope of a dimensionless point on a curved slope possible. And doing that seems to be the major hurdle to opening up things for a calculus. But almost immediatly we are taught to make derivatives by a totally different method -- reducing the exponant by one, etc. So that, for exmple 2x^3 becomes 6x^2 etc. Well, that is of course lots easier. But where is the connection? How does one get from that magical limits equation (which shows how it really does work) to the the cheap trick? (Not that I am necessarily against cheap tricks.)

 

[Opalg]

Here's a question. This formula seems to be the keystone of calculus.

View attachment 644That seems to be what the calculus books say, and it makes sense to me, as a rank beginner. This equation is what makes the seeming magic of defining the slope of a dimensionless point on a curved slope possible. And doing that seems to be the major hurdle to opening up things for a calculus. But almost immediatly we are taught to make derivatives by a totally different method -- reducing the exponant by one, etc. So that, for exmple 2x^3 becomes 6x^2 etc. Well, that is of course lots easier. But where is the connection? How does one get from that magical limits equation (which shows how it really does work) to the the cheap trick? (Not that I am necessarily against cheap tricks.)

That is an excellent question, and shows that you are thinking in the right way for a mathematician! Too often, calculus is taught as though it is just a box of tricks, when in fact these "tricks" follow logically from the definitions.

Take your example of the function $f(x) = 2x^3$. If you apply the "keystone" formula, then the derivative of this function at the point $x=a$ is $$f'(a) = \lim_{h\to 0}\frac{f(a+h)-f(a)}{h} = \lim_{h\to 0}\frac{2(a+h)^3 - 2a^3}{h}.$$ Now you have to do a bit of algebra, to calculate that $2(a+h)^3 = 2a^3 + 6a^2h+6ah^2+h^3.$ It follows that $$\frac{2(a+h)^3 - 2a^3}h = \frac{6a^2h + 6ah^2 + h^3}h = 6a^2 + 6ah + h^2.$$ Then as $h\to0$, the last two terms in that expression go to $0$ and you see that $$\lim_{h\to0}6a^2 + 6ah + h^2 = 6a^2.$$ So the derivative of $6x^3$ at $x=a$ is $6a^2.$ This is usually written in an abbreviated form by saying that the derivative of $6x^3$ is $6x^2.$

 

[zzephod]

Tim Gower has something to say about this on his blog:

http://gowers.wordpress.com/2012/11/20/what-maths-a-level-doesnt-necessarily-give-you/

.

Moderator edit: This interesting blog sparked interest in another discussion, which I have moved http://www.mathhelpboards.com/f9/counter-intuitive-phenomena-math-physics-3637/ in order to keep this discussion on topic.

 

[jedson303]

OK, Opalg. I see how that works. It does connect. Thanks.

 

[Prove It]

Here's a question. This formula seems to be the keystone of calculus.

View attachment 644That seems to be what the calculus books say, and it makes sense to me, as a rank beginner. This equation is what makes the seeming magic of defining the slope of a dimensionless point on a curved slope possible. And doing that seems to be the major hurdle to opening up things for a calculus. But almost immediatly we are taught to make derivatives by a totally different method -- reducing the exponant by one, etc. So that, for exmple 2x^3 becomes 6x^2 etc. Well, that is of course lots easier. But where is the connection? How does one get from that magical limits equation (which shows how it really does work) to the the cheap trick? (Not that I am necessarily against cheap tricks.)

Let's say we want to find the derivative of $\displaystyle \begin{align*} f(x) = k\,x^n \end{align*}$, where k is a constant and n is a positive integer. Then we have

\displaystyle \begin{align*} f'(x) &= \lim_{h \to 0}\frac{f(x + h) - f(x)}{h} \\ &= \lim_{h \to 0}\frac{k\left( x+ h \right)^n - k\,x^n}{h} \\ &= \lim_{h \to 0}\frac{ k \sum_{r = 0}^n{ \left[ {n\choose{r}} x^{n-r} h^r \right] } - k\,x^n }{h} \\ &= \lim_{h \to 0}\frac{k\,x^n + k\sum_{r = 1}^n{\left[ {n\choose{r}} x^{n-r} h^r \right] } - k\,x^n }{h} \\ &= \lim_{h \to 0}\frac{ k\sum_{r = 1}^n{ \left[ {n\choose{r}} x^{n-r} h^r \right] } }{h} \\ &= \lim_{h \to 0} k\sum_{r = 0}^n{ \left[ {n\choose{r}} x^{n-r} h^{r-1} \right] } \\ &= \lim_{h \to 0}\left\{ n\,k\,x^{n-1} + k\sum_{r = 2}^n{ \left[ {n\choose{r}}x^{n-r}h^{r-1} \right] } \right\} \\ &= n\,k\,x^{n-1} \end{align*}

The result easily extends to other types of powers, but might need knowledge of the chain rule or other rules.

 

[jedson303]

I'm going back to review the Infinite sequences and summation notations, to better follow all the calculations. But I can see what you are doing. This way you don't just prove the validity of a function with a specific value, but show why this holds for all values of that function. How, in other words, it can be deduced from the "keystone" formula, and then used as a valid rule. Which is what I asked. Interesting.

Jedson

 

[Prove It]

I'm going back to review the Infinite sequences and summation notations, to better follow all the calculations. But I can see what you are doing. This way you don't just prove the validity of a function with a specific value, but show why this holds for all values of that function. How, in other words, it can be deduced from the "keystone" formula, and then used as a valid rule. Which is what I asked. Interesting.

Jedson

It's not an infinite sequence, or an infinite series for that matter. It's a particular finite series called the Binomial Theorem or Binomial Expansion.

 

[jedson303]

I guess I didn't understand what you were doing. I'll review the binomal therum and see if I can follow the computation. Then try to understand what it actually means.
jedson