Difference Quotient

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Jason76

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Look right:confused:

\(\displaystyle f(x) = x^{2} - 6x\)

Find derivative using difference quotient:

Note: Brackets used to make a space, to make more readable.

\(\displaystyle \lim h \to 0 [\dfrac{f(x + h) - f(x)}{h}]\) - Remember to move constants outside of parenthesis when dealing with \(\displaystyle (x + h)\), and exponents outside also.

\(\displaystyle \lim h \to 0 [\dfrac{(x + h)^{2} - 6(x + h) - (x^{2} - 6x)}{h}]\)

\(\displaystyle \lim h \to 0 [\dfrac{(x + h)(x + h) - 6x - 6h - (x^{2} - 6x)}{h}]\)

\(\displaystyle \lim h \to 0 [\dfrac{x^{2} + 2xh + h^{2} - 6x - 6h - x^{2} + 6x}{h}]\)

\(\displaystyle \lim h \to 0 [\dfrac{2xh + h^{2} - 6x - 6h + 6x}{h}]\)

\(\displaystyle \lim h \to 0 [\dfrac{2xh + h^{2} - 6h}{h}]\)

\(\displaystyle \lim h \to 0 [2xh/h + h^{2}/h - 6h/h]\)

\(\displaystyle \lim h \to 0 [2x + h - 6]\)

\(\displaystyle \lim h \to 0 [2x + (0) - 6] = 2x - 6\)

\(\displaystyle f ' (x) = x^{2} - 6x = 2x - 6\)
 
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Jason76 said:
Needed to put in limit notation and substitute in \(\displaystyle 0\) for \(\displaystyle h\) to be really correct.
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Yeah, that's a great way to lose points on an exam, as I'm sure your instructor would want everything to be properly written.
 
Using the power rule (much simpler) :D

\(\displaystyle f'(x) = nx^{n - 1}\)

\(\displaystyle f'(x) = x^{2} - 6x\)

\(\displaystyle f'(x) = (2)x^{2 - 1} - (1)6x^{1 - 1}\)

\(\displaystyle f'(x) =(2)x^{1} - (1)6x^{0}\)

\(\displaystyle f'(x) = 2x - 6\)
 
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Jason76 said:
Using the power rule (much simpler) :D

\(\displaystyle f'(x) = x^{2} - 6x\)

\(\displaystyle (2)f'(x) = (2)x^{2 - 1} - (1)6x^{1 - 1}\)

\(\displaystyle (2)f'(x) =(2)x^{1} - (1)6x^{0}\)

\(\displaystyle f'(x) = 2x - 6\)
Click to expand...

Well, of course, but in Calculus that's usually how it works. Learn it the hard way first, then find an infinitely easier way to solve the problem. Which book does your class use? Stewart's?
 
Kaitlyn25 said:
Well, of course, but in Calculus that's usually how it works. Learn it the hard way first, then find an infinitely easier way to solve the problem. Which book does your class use? Stewart's?
Click to expand...

I know, just joking around. Yes. Stewarts.
 
Kaitlyn25 said:
Well, of course, but in Calculus that's usually how it works. Learn it the hard way first, then find an infinitely easier way to solve the problem. Which book does your class use? Stewart's?
Click to expand...
I agree that you could first derive the power rule and then apply it. Where is your derivation of the power rule?
 
Jason76 said:
Don't know that one. Just know how to use the power rule.
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Well, this exercise shows you why someone made the effort to prove that the power rule was true.

In essence you just showed that the power rule is true for the powers of 2 and 1 in this exercise.

Now I am not showing off here. I could derive the power rule for integer powers very easily. I would have at least difficulty doing so for powers that were multiplicative inverses of integers. And I doubt I could do it at all for irrational powers. A lot of what you do in a beginning calculus course is to learn how to use rules that are not really proved. That is, the course is primarily about memorizing and applying rules with just a bit of justification for the rules thrown in. (At least that is what calculus was when I learned it way back when.) The justification is hard. It took mathematicians about two hundred years to do it.
 
The derivation of the power rule is all fun and games until the powers become irrational!
 
or you could write:

\(\displaystyle \displaystyle f'(x) \ = \ \frac{d}{dx}(x^2 \ - \ 6*x) \ = \ 2*x - 6\)
 
Okay, I will "show off"!

There are two ways typically used to prove the "power rule". One is to use the "binomial theorem":
\(\displaystyle (x+ y)^n= \sum_{i= 0}^n\begin{pmatrix}n \\ i \end{pmatrix}x^{n-i}y^i\)

If \(\displaystyle f(x)= x^n\) then \(\displaystyle f(x+h)= (x+ h)^n= \sum_{i=0}^n \begin{pmatrix}n \\ i\end{pmatrix}x^{n-i}h^i\) so that \(\displaystyle f(x+h)- f(x)= \sum_{i= 1}^n\begin{pmatrix}n \\ i\end{pmatrix}x^{n-i}h^i\)
\(\displaystyle \dfrac{f(x+h)- f(x)}{h}= \dfrac{\sum_{i= 1}^n\begin{pmatrix}n \\ i\end{pmatrix}x^{n-i}h^i}{h}= \sum_{i= 1}^n\begin{pmatrix}n \\ i\end{pmatrix}x^{n-i}h^{i- 1}\)

Taking the limit as h goes to 0 every term except the \(\displaystyle h^0= h^{1- 1}\) term, with i= 1, is 0 so
\(\displaystyle (x^n)'= \lim_{h\to 0}\dfrac{f(x+h)- f(x)}{h}= \begin{pmatrix}n \\ 1 \end{pmatrix}x^{n- 1}= nx^{n-1}\).

The other way, once we have proven that the derivative of f(x)= x is f'(x)= 1 and the product rule, is to use "proof by induction".

Base case: n= 1. \(\displaystyle f(x)=x^1= x\) so \(\displaystyle f'(x)= 1= 1(x^0)= 1(x^{1- 1})\).

Assume that, for some k, if \(\displaystyle f(x)= x^k\) then \(\displaystyle f'(x)= kx^{k-1}\). Then if \(\displaystyle f(x)= x^{k+ 1}= x(x^k)\), by the product rule, \(\displaystyle f'(x)= (x)' x^k+ (x)(x^k)'= (1)x^k+ (x)(kx^{k-1})= x^k+ kx^k= (k+1)x^k\).
 
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But the above proof assumes n to be an integer
excludes fractional n (which includes irrational n)
 
JeffM said:
Well, this exercise shows you why someone made the effort to prove that the power rule was true.

In essence you just showed that the power rule is true for the powers of 2 and 1 in this exercise.

Now I am not showing off here. I could derive the power rule for integer powers very easily. I would have at least difficulty doing so for powers that were multiplicative inverses of integers. And I doubt I could do it at all for irrational powers. A lot of what you do in a beginning calculus course is to learn how to use rules that are not really proved. That is, the course is primarily about memorizing and applying rules with just a bit of justification for the rules thrown in. (At least that is what calculus was when I learned it way back when.) The justification is hard. It took mathematicians about two hundred years to do it.
Click to expand...

Perhaps it would be a masters degree or PHD thesis.
 
Subhotosh Khan said:
But the above proof assumes n to be an integer
excludes fractional n (which includes irrational n)
Click to expand...
Yes, for n non-integer, I would wait until after I had proved that \(\displaystyle \dfrac{dln(x)}{dx}= \dfrac{1}{x}\) and the chain rule and then say, "if \(\displaystyle f(x)= x^n\) then ln(f)= n ln(x) so that \(\displaystyle \dfrac{1}{f(x)}\dfrac{df}{dx}= \dfrac{n}{x}\) so that \(\displaystyle \dfrac{df}{dx}= n\dfrac{x^n}{x}= nx^{n-1}\)".
 
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