2020-04-07 22:13:52 (edited by Ethin 2020-04-07 22:24:34)

So, I'm getting confused. I was taking my exam for derivatives. I can't bring examples to follow along with (which I work best with) so I was forced to just take down notes. The problem is that even though I had the various rules written down from OpenStax, I got stuck when attempting to figure out what rule applied to what problem. For example, if the exam gave me a problem like 12x^2-19x/14x, I got confused because most of the rules for derivatives require two functions, not one, and I wasn't sure how to find g(x) even if I had the original equation, which I automatically classified as f(x). Thoughts?
Edit: forgot to add that the chain rule seems, based on its definition, to be convoluted as all hell. OpenStax identifies the chain rule as: "let f and g be functions. For all x in the domain of g for which g is differentiable at x and f is differentiable at g(x), the derivative of the composite function h(x)=(fºg)(x)=f(g(x)) is given by h'(x)=f'(g(x))g'(x); alternatively, if y is a function of u and u is a function of x then dy/dx=dy/du(du/dx)." Wikipedia indicates the same but in a much more verbose form, and it makes a bit more sense that way, but I'm still stuck on how to determine g(x).

"On two occasions I have been asked [by members of Parliament!]: 'Pray, Mr. Babbage, if you put into the machine wrong figures, will the right answers come out ?' I am not able rightly to apprehend the kind of confusion of ideas that could provoke such a question."    — Charles Babbage.
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2020-04-07 22:53:49

@26, is your example meant to be taken as:

(12x^2-19x)/14x

or

12x^2-(19x/14x)

Given you're being exposed to the rules for differentiation, it would make sense to me that the first interpretation is correct since it would require using the quotient rule to solve and not by splitting the derivative up into two derivatives by subtraction. Just want to make sure.
Also, have you covered derivatives of trig functions? That will help you understand this better.

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2020-04-07 23:40:36

@27, we have, a bit, but I haven't really practiced that that much and don't fully understand it (my tutor tried to explain it but it was kinda confusing).

"On two occasions I have been asked [by members of Parliament!]: 'Pray, Mr. Babbage, if you put into the machine wrong figures, will the right answers come out ?' I am not able rightly to apprehend the kind of confusion of ideas that could provoke such a question."    — Charles Babbage.
My Github

2020-04-07 23:55:52

@28, derivatives of trig functions are actually just things to commit to memory. There are some things to notice, like the ones starting with "c" have a negative. cos, cosecant, cotangent all have negative derivatives.
p.s.: let us know which way to read your example in your second to last post.

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2020-04-08 00:22:24

leibylucw wrote:

Some blind students take the "spend 6-8 hours a week writing everything down in LaTeX" approach in their math classes which is ridiculously wasteful to me, but as long as they're happy in doing so, more power to those folks.

I feel triggered! lol Just kidding, kind of! I do get this though, but sometimes you just got to do what it takes. And I didn't have access to Braille, and quite frankly didn't feel like advocating for it. I wrote a program to convert plain text to various formats, but still was somewhat time consuming. As already said, we're just going to be slower than our sighted peers, even if we use Braille or some fancy programs. At the end of the day, the aim is to get that little piece of paper that says we are competent enough to stick by an unforgiving subject such as Math, Physics, or engineering.

2020-04-08 00:55:25

@30, ugh, you LaTeX nerds
tongue
But seriously, committing to such a feat in classes that are already challenging enough is very admirable/impressive. As much as I strongly suggest students take a route similar to what I've done/am doing (assuming they learn in similar ways), I do recognize there are huge disparities that are not as easily solvable. The hardships we face are systemic and far more difficult to overcome. If it were to take me that much effort to complete even one class, I would have to sit down and really think if A) I'm the kind of person who has the self discipline to do so and B) is it worth doing so?
All that to say kudos to you and the others I know who have sat down and put in those long hours writing your work down and grinding to get the job done. If you can turn that into a degree, that must mean it's effective.
smile

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2020-04-08 02:27:08

@29, it was the first interpretation.

"On two occasions I have been asked [by members of Parliament!]: 'Pray, Mr. Babbage, if you put into the machine wrong figures, will the right answers come out ?' I am not able rightly to apprehend the kind of confusion of ideas that could provoke such a question."    — Charles Babbage.
My Github

2020-04-08 18:46:42 (edited by leibylucw 2020-04-08 21:51:32)

@32, here's my best attempt at explaining the concepts you mentioned:

As far as the problem you presented, you would use the quotient rule to solve. The rule states:

The derivative of a rational function is calculated by FPrime*G-F*GPrime all divided by G^2,
where F(X) = the numerator and G(X) = the denominator and you are [NOT] dividing by 0.

Knowing this, you can solve for both FPrime and GPrime by deriving the numerator and denominator. At that point, it's just a matter of plugging things and doing the algebra. Just to make sure we both are getting the same answers, I contend that:

FPrime = d/dx(12x^2-19x) = 24x-19
GPrime = d/dx(14x) = 14

You now have all of what you need to plug in and solve using the quotient rule. It's just algebraic work.

Regarding the chain rule, it's actually far more simple than the definition you provided is leading you to believe. All it states is:

The derivative of F(G(X)), or, in plain terms, a function composed of another function, is equal to the derivative of F(X), or the outside function, times the derivative of G(X), or in the inside function.

Let's say we want to derive (4x^3+2x)^2. Note that we are raising the quantity of 4x^3+2x to the second power, as delineated by the parentheses. Whatever is inside of the parentheses is its own function, but we are raising that whole thing to an exponent, so we split this up into two functions. The chain rule is most apparent through exponents, so this is probably the most common type of problem you'll see next to trig derivatives. Now, we represent what I just stated as:

F(X) = ( G(X) )^2, where G(X) is just another function.
G(X) = (4x^3+2x)

It really helps to force parentheses around things to help make it clear which terms are part of a single function. In this case, the stuff inside the parentheses surrounding G(X) will eventually be raised to a second power, meaning whatever value G(X) produces will be squared.
Now, and this is also extremely important, when you are deriving F(X), you do not consider anything inside of the parentheses. F(X) was defined as a placeholder that squares whatever is inside of it. In other words, don't even think about G(X) yet, we're just focused on ( ... )^2, where ... is just stuff inside of F(X).
To make this even more apparent, let's just say that ( ... ) = u. Then we see that we are deriving u^2.
Just like you would do for x^2, deriving u^2 gives us 2u. Plugging ( ... ) back in for u, we now have 2( ... ). Plugging G(X) back in for ..., we now see that the derivative of the outside function is 2(4x^3+2). That is precisely how you derive the outside function, or F(X).

To solve for the derivative of the inside function, it's actually simple. Recall that we let G(X) be equal to 4x^3+2x. Deriving that gives us 12x^2+2.
Finally, we multiply both of these things together. So we'd have:

2(4x^3+2x)
*
(12x^2+2)

to give us:

2*(4x^3+2x0*(12x^2+2)

If you really wanted to, you could distribute all of that out, but it's needless algebra at this point. No use in simplifying.

I can understand the confusion over how you define F and G. Try to remember that if you can place parentheses around terms which then are being altered by exponents, everything outside of the parentheses is your F. If we had multiplied all of this by 2, it would've looked like:

2(4x^3+2x)^2

so we'd be deriving 2( ... )^2, or 2u^2.

You'll also probably find it easier to understand all of this by using the substitutions like I did. It's really deriving the outer function people don't quite understand, but if you're not getting it, try replacing things with names that look similar to variables so it becomes more apparent when you're differentiating.

Let me know if any of that made sense.

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2020-04-08 21:41:32

@33, that did! Thanks for that!

"On two occasions I have been asked [by members of Parliament!]: 'Pray, Mr. Babbage, if you put into the machine wrong figures, will the right answers come out ?' I am not able rightly to apprehend the kind of confusion of ideas that could provoke such a question."    — Charles Babbage.
My Github

2020-04-10 03:40:47

Anytime! Feel free to ask any questions you have. I can try to help, and I'm sure others will chime in too.

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2020-04-16 03:01:31

I am curious (and trying to keep the topic alive a little bit), how many are utilizing LaTeX and MathML? Are any Braille readers using BRF's?

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2020-04-16 04:39:22

So for math I use the KeyMath map conversion utility that's inside the BrailleNote Touch Plus, which converts any Braille math symbols into print ones (as long as I write them correctly).

A winner is you!
—Urban Champion

2020-04-16 06:28:56

@37, interesting. Are you using nemeth or UEB? I think you're one of the calc folks, so that'd be neat to hear how the conversion is going for you.

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2020-05-29 13:17:46

I know this is a bit old, but wanted to bring it back up to see how the rest of the semester/school year went for everybody.
How did the rest of your math courses go? Did the accessibility issues get ironed out in light of COVID19's forcing classes to be online?
My Calc 2 final went pretty well. It was a bit strange doing a math final over the phone, but it worked really well and I got a good grade.
I'm fortunately done with the math track of my major, so here's to just taking CS classes!
big_smile
I hope others have been fortunate as well. This whole being blind thing and taking STEM courses is such a pain.

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2020-05-29 15:08:35

Well, math ended up being surprisingly easy. Florida's governor canceled all standardized testing for K-12 students, which meant that all classes that relied on an end of course exam would be graded as if the tests didn't exist. But with my teacher, he would call each student weekly and he would work on the homework with us one on one. We would be sent braille handouts weekly, type our answers in a word doc, e-mail our answers to him, and call it a day

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PSN ID, Rich_Beardsley

2020-05-29 15:38:42

Standardized testing is blah, especially in certain areas where you're being assessed on concepts you haven't explicitly seen in months or possibly years. We have our own version of it in PA, and it always bugged everybody. I escaped that misery just as it began my senior year.
Glad to hear your teacher was very accommodating! These are very tryiing times, so it's refreshing to hear educators are stepping up and really giving it their all to afford their students the opportunity to learn and demonstrate their knowledge.

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