mersenneforum.org Introductory Calculus Discussion Thread
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2018-10-16, 23:49   #56
jvang
veganjoy

"Joey"
Nov 2015
Middle of Nowhere,AR

1101111002 Posts

Quote:
 Originally Posted by VBCurtis On your last line, couldn't you factor an x out of the top instead and make the same argument about an expression times infinity? What would that limit be? Your last line has 0 times an unknown quantity. What if that unknown quantity goes to inifinity while your known term goes to zero? Note that, even if you know what infinity times zero is (you don't), you have a quantity that might be "going to infinity" multiplied by a quantity that is definitely "going to zero". The problem is that you don't know whether one is growing faster than the other is shrinking, so it's not so wise to break up limits in the way you did (unless you can demonstrate that the "unknown" part goes to a constant, which in this example it does via the divide-by-biggest-power-on-bottom trick you already discussed).
Ok...I think I get what you're saying. I posed it to my teacher and she was just like "yeah, sure"

I think my problem was treating the first limit term as a constant, since I hadn't considered the concept of infinity times zero. Is my understanding of limit laws incorrect/misguided?

 2018-10-17, 00:53 #57 VBCurtis     "Curtis" Feb 2005 Riverside, CA 23×3×211 Posts I think your understanding of the laws is fine; your flaw was in not considering whether the first limit was a constant or infinite. If you demonstrate the first limit is a constant (in your example, it indeed is), then your use of the laws is correct and your answer is supported by your work.
2018-11-23, 16:10   #58
jvang
veganjoy

"Joey"
Nov 2015
Middle of Nowhere,AR

22×3×37 Posts

Quote:
 Originally Posted by jvang I'm under the impression that we aren't actually going to learn the formal epsilon-delta definition of a limit, given that we're jumping to the squeeze theorem without a proof of how it or limits work
We started on derivatives a couple of weeks ago, still no formal definition of limits. We’ve done a ton of work, but covered very little in actual concepts. I can’t really remember what we have covered!

One thing I remember is the power laws for derivatives. The derivative of a function is equal to the sum of each term modified as such: $\frac{d(x)}{x} 2x^2 = 2*2x^{2-1} = 4x$You multiply the coefficient by the exponent, then subtract 1 from the exponent. Another example:$\frac{d(x)}{x} x^3 - 4x + 2 = 3x^2 - 4$We haven’t covered the proper notation, is d(x)/x correct?

2018-11-24, 06:19   #59
ATH
Einyen

Dec 2003
Denmark

31×103 Posts

Quote:
 Originally Posted by jvang Another example:$\frac{d(x)}{x} x^3 - 4x + 2 = 3x^2 - 4$We haven’t covered the proper notation, is d(x)/x correct?
The notation is:

$\frac{d}{dx}(x^3 - 4x + 2) = 3x^2 - 4$

or alternately

$\frac{d(x^3 - 4x + 2)}{dx}= 3x^2 - 4$

2018-11-24, 10:03   #60
Nick

Dec 2012
The Netherlands

33368 Posts

Quote:
 Originally Posted by jvang One thing I remember is the power laws for derivatives.
Did they make it clear why it works like that?

Take your example of 2x².
If we draw a graph for this function, we get lots of points of the form (x,y) where x is any number and y=2x².
Fix the value of x and take another value a bit further on, say x+d for some small number d.
The 2 corresponding points on the graph then have coordinates (x,2x²) and (x+d,2(x+d)²).
How steep is the line that passes through these 2 points?
Well, the difference between the 2nd coordinates is

$2(x+d)^2-2x^2=2(x^2+2dx+d^2)-2x^2=4dx+2d^2$
while the difference between the 1st coordinates is $$(x+d)-x=d$$,
so the ratio between these 2 differences is
$\frac{4dx+2d^2}{d}=4x+2d$
If we now take the limit as d tends to 0, this ratio becomes 4x.
That's why we have $$\frac{d}{dx}2x^2=4x$$.

2018-11-24, 15:02   #61
jvang
veganjoy

"Joey"
Nov 2015
Middle of Nowhere,AR

22·3·37 Posts

Quote:
 Originally Posted by Nick Did they make it clear why it works like that? Take your example of 2x². If we draw a graph for this function, we get lots of points of the form (x,y) where x is any number and y=2x². Fix the value of x and take another value a bit further on, say x+d for some small number d. The 2 corresponding points on the graph then have coordinates (x,2x²) and (x+d,2(x+d)²). How steep is the line that passes through these 2 points? Well, the difference between the 2nd coordinates is $2(x+d)^2-2x^2=2(x^2+2dx+d^2)-2x^2=4dx+2d^2$ while the difference between the 1st coordinates is $$(x+d)-x=d$$, so the ratio between these 2 differences is $\frac{4dx+2d^2}{d}=4x+2d$ If we now take the limit as d tends to 0, this ratio becomes 4x. That's why we have $$\frac{d}{dx}2x^2=4x$$.
In a much more dumbed-down way, we did do a silly little activity that attempted to demonstrate how that works

To solve for the derivative without the power rule, our teacher gave us a “definition” to plug numbers into:$\dfrac{f(x+h)-f(x)}{h}$Which looks similar to what you posted. Then we solve it as a limit as $$h$$ approaches 0. It’s not very complicated; the numerator seems to always simplify to terms that all divide by $$h$$. For instance, the derivative of $$3x-7$$:$\dfrac{3(x+h)-7-(3x-7)}{h}=\dfrac{3x+3h-7-3x+7}{h}=\dfrac{3h}{h}=3$It does get a little messy with exponents, with over a dozen terms sometimes, but the numerator always ends up divisible by $$h$$.

2018-11-25, 09:17   #62
Nick

Dec 2012
The Netherlands

175810 Posts

Quote:
 Originally Posted by jvang It does get a little messy with exponents, with over a dozen terms sometimes, but the numerator always ends up divisible by $$h$$.
Take any number a and a positive integer n and define f(x)=axⁿ.
For any numbers x and h, by the binomial theorem we have
$(x+h)^n=x^n+nx^{n-1}h+\frac{n(n-1)}{2!}x^{n-2}h^2+\frac{n(n-1)(n-2)}{3!}x^{n-3}h^3+\ldots +nxh^{n-1}+h^n$
so
$\begin{eqnarray*} \frac{f(x+h)-f(x)}{h} & = & \frac{a(x+h)^n-ax^n}{h} \\ & = & \frac{a(x^n+nx^{n-1}h+\frac{n(n-1)}{2!}x^{n-2}h^2+\ldots +nxh^{n-1}+h^n)-ax^n}{h} \\ & = & \frac{a(nx^{n-1}h+\frac{n(n-1)}{2!}x^{n-2}h^2+\ldots +nxh^{n-1}+h^n)}{h} \\ & = & nax^{n-1}+\frac{n(n-1)}{2!}ax^{n-2}h+\ldots +naxh^{n-2}+ah^{n-1}. \end{eqnarray*}$
If we take the limit as h tends to 0 then all terms except the first disappear, leaving us with $$nax^{n-1}$$.
This proves the power law.

 2018-11-26, 23:48 #63 jvang veganjoy     "Joey" Nov 2015 Middle of Nowhere,AR 22×3×37 Posts We just went over 2 new derivative rules, involving products and quotients. Product rule:$f(x)=g(x)h(x), \ f'(x)=g'(x)h(x)+g(x)h'(x)$Our teacher said that using a tickmark/apostrophe/prime symbol denotes taking the derivative of a function. Quotient rule:$f(x)=\dfrac{g(x)}{h(x)}, \ f'(x)=\dfrac{g'(x)h(x)-g(x)h'(x)}{g(x)^2}$We haven’t covered the properties of functions that tell us whether we can take the derivative of a function, other than continuity. Thus, I can’t really make much sense of these rules, and will just have to memorize and regurgitate them on exams. Nick's demonstration involving the binomial theorem was really helpful and neat. For instance, I was wondering whether you could use the product rule on polynomials like these: $$(3x^2-2)(x^2+5)^{-1}$$, but the case wasn’t covered in our book and I didn’t have the time to ask my teacher. I know that the derivative of the second term would be $$-2x^{-3}$$, but I’m not sure where I could go from there...
 2018-11-27, 04:01 #64 VBCurtis     "Curtis" Feb 2005 Riverside, CA 23×3×211 Posts The product rule will work on the example you suggested, once you learn the Chain Rule (coming next). The chain rule will allow you to take the derivative of an expression to a power, such as the second term in your example. The chain rule: d/dx [f(g(x)] = f'(g(x)) * g'(x). In casual English: take the derivative of the outside function, leaving the inside untouched; then multiply by the derivative of the inside function. So, the derivative of$(x^2 +5) ^{-1} = -1 (x^2 +5)^{-2} (2x)$.
2018-11-27, 05:24   #65
masser

Jul 2003
Behind BB

25×5×11 Posts

Quote:
 Originally Posted by jvang We just went over 2 new derivative rules, involving products and quotients. Product rule:$f(x)=g(x)h(x), \ f'(x)=g'(x)h(x)+g(x)h'(x)$Our teacher said that using a tickmark/apostrophe/prime symbol denotes taking the derivative of a function. Quotient rule:$f(x)=\dfrac{g(x)}{h(x)}, \ f'(x)=\dfrac{g'(x)h(x)-g(x)h'(x)}{g(x)^2}$We haven’t covered the properties of functions that tell us whether we can take the derivative of a function, other than continuity. Thus, I can’t really make much sense of these rules, and will just have to memorize and regurgitate them on exams. Nick's demonstration involving the binomial theorem was really helpful and neat. For instance, I was wondering whether you could use the product rule on polynomials like these: $$(3x^2-2)(x^2+5)^{-1}$$, but the case wasn’t covered in our book and I didn’t have the time to ask my teacher. I know that the derivative of the second term would be $$-2x^{-3}$$, but I’m not sure where I could go from there...
Can you show (with some algebraic manipulations) that :

f(x) = (3x^2-2)(x^2+5)^{-1} = 3 - 17(x^2+5)^{-1}

Then, using the chain rule idea, you should get

f'(x) = 34x(x^2+5)^{-2}

[This way is a little more "fun" than just applying the rule for differentiating fractions g(x)/h(x), where g(x) = 3x^2-2 and h(x) = x^2+5. It should give the same answer...]

Last fiddled with by masser on 2018-11-27 at 05:29

2018-11-27, 23:07   #66
Dr Sardonicus

Feb 2017
Nowhere

24×11×29 Posts

Quote:
 Originally Posted by masser Can you show (with some algebraic manipulations) that : f(x) = (3x^2-2)(x^2+5)^{-1} = 3 - 17(x^2+5)^{-1} Then, using the chain rule idea, you should get f'(x) = 34x(x^2+5)^{-2} [This way is a little more "fun" than just applying the rule for differentiating fractions g(x)/h(x), where g(x) = 3x^2-2 and h(x) = x^2+5. It should give the same answer...]
You can always (in theory) express a rational function as a polynomial plus a "proper fraction" whose numerator has lower degree than the denominator, using polynomial division with quotient and remainder. This can indeed simplify the algebra when taking the derivative.

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