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Cheap tricks

jedson303

New member
Feb 23, 2013
12
Here's a question. This formula seems to be the keystone of calculus.

limit.jpgThat 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 dimentionless 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.)
 

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Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,702
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 dimentionless 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 $6x^2.$
 
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zzephod

Well-known member
Feb 3, 2013
134
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jedson303

New member
Feb 23, 2013
12
OK, Opalg. I see how that works. It does connect. Thanks.
 
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Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
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 dimentionless 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

[tex]\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*}[/tex]

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

jedson303

New member
Feb 23, 2013
12
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

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
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

New member
Feb 23, 2013
12
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