/sci/ - Science & Math

[Curious Calculus Student No.5136471]

Hi there.

I need help learning Calculus - more specifically, derivatives.
I know this is very low level math, but please bare with me.

My teacher only handed out to us a sheet with the Derivatives Rules, with no examples. Accompanying the Rules handout was ~17 practice problems and a problem set to be handed in tomorrow (4% of our final mark).

I'm not here to ask for anyone to do my homework for me, rather, where I can learn (examples are a must) to effectively use the derivatives rules to solve questions such as:

1. y * cos(x) - csc(y) = y^(5-(x^2)), Determine D(x) y
2. 3 * (y * sin(x)) ^3 = y * sin(5x) - x * cos(3y), Determine D(y) x
3. (y-x)/5 = (2x)^y - 3(log(base 4)(xy)), Determine D(x) y
4. (4x)/y = 3^(y*tan^2x) - y^(3-x) - 1, Determine dy/dx

Thanks in advance.

[Curious Calculus Student No.5136482]

>>5136471
I should go on to say that I'm also having problems with the practice problems. Though my final goal for tonight is to be able to at least answer the four above-mentioned questions.

[Curious Calculus Student No.5136495]

Bumping with math reference sheets.

[Anonymous No.5136516]

>>5136471

They are some ugly looking questions.

You will need to know implicit differentiation for all of them (power and chain rules will help also)

Check out Khan Academy to get a rough idea on what to do, and if all else fails use Wolfram Alpha, get the answers hand it in, then learn the material (If you get examined and cannot perform what you did in assignments, marks can be rescinded).

[Anonymous No.5136521]

What exactly does the D operator there stand for, since I assume D(x) doesn't mean d/dx or something since you wrote dy/dx for the bottom one.

Either way, I think I can help you with the bottom one.

You can write y as a function, call it f(x), so replace every y with f(x) in that equation. Now you just differentiate both sides of the equation with respect to x, and you'll get an equation with f(x) and f'(x) in it. Now replace every f(x) with y again, and solve for f'(x). It looks pretty nasty for that equation though, so it could be a lot of work.

[Curious Calculus Student No.5136525]

>>5136516
Will give it a try.
Although, for some reason, I'm having a problem with the most basic questions, such as:

y=[(2x - 3)^3 - (3x^2)] / (x-1), Determine D(x) y

[Anonymous No.5136530]

>>5136521
That is the derivative with respect to x.

I'll hunt down some examples for you.

[Curious Calculus Student No.5136531]

>>5136521
D(x) y means: "the derivative of y with respect to x"
Sorry for the misunderstanding, as (x) in D(x) y is supposed to be subscripted.

[Anonymous No.5136543]

>>5136521
http://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/implicitdiffdirectory/ImplicitDiff.html

Go down to problems and see how they are solved.

But, if you don't have an intuitive understanding of calculus, and you're being given these questions -- you have either paid little to no attention and it's your fault, or your teacher is a failure (be honest to yourself).

[Anonymous No.5136549]

>>5136531
So why not write dy/dx like in the 4th question, I feel like I'm perhaps misunderstanding the question. But if that's it I would propose the same method as I mentioned for the 4th question.

>>5136525
So for this example you would say x = f(y), and the equation becomes. y = [(2 f(y)-3)^3-(3 f(y)^2)]/(f(y)-1). Take the derivative of both sides. So you'll get:
<div class="math"> 1 = \frac{(-27+f[y] (78+f[y] (-63+16 f[y]))) f'[y]}{(-1+f[y])^2}</div>

And now you just say f[y] = x and then solve for f'[y]. And you should get:
<div class="math">f'[y] = \frac{(-1+x)^2}{-27+78 x-63 x^2+16 x^3}</div>

(I hope I didn't mess up my LaTex code)

[Anonymous No.5136578]

http://vocaroo.com/i/s1hPPxSly2bF

[Curious Calculus Student No.5136589]

>>5136549
b-but...

[Anonymous No.5136607]

>>5136578
If you want to play the cool guy, then at least don't talk total crap. All you did was showing off your ignorance of babby math. Why are you doing this anyway? Guessing from your voice you're way too old to deal with such simple highschool shit.

[Anonymous No.5136620]

>>5136589
Oh I'm sorry I did dy/dx, that was stupid haha.

[Anonymous No.5136626]

>>5136620
>>5136589
Ugh I mean dx/dy.

[Anonymous No.5136638]

>>5136607

>don't talk crap
>talks crap

woah it's like I really am in high school

[Anonymous No.5136661]

Alright I'll try to explain it better with a simple example. Let's say we have the equation <span class="math"> x^4 y + 3y^3 = x + y[/spoiler] and we want to know <span class="math"> \frac{dy}{dx}[/spoiler]. We can make the subsitution y = f(x), so the equation becomes <span class="math"> x^4 f(x) + 3 f(x)^3 = x + f(x)[/spoiler]. Now we take the derivative of both sides and substitute back:
<div class="math"> \frac{d}{dx}(x^4 f(x) + 3 f(x)^3 = x + f(x))</div> so <div class="math"> 4 x^3 f(x)+x^4 f'(x)+9 f(x)^2 f'(x)=1+f'(x)</div> so <div class="math">4 x^3 y+x^4 f'(x)+9 y^2 f'(x)=1+f'(x)</div> Now you solve for f'(x) so we get <div class="math"> f'(x)= \frac{1-4 x^3 y}{-1+x^4+9 y^2} = \frac{dy}{dx}</div> Now use that method for the other questions as well, it'll be a lot tougher but the idea is the same.

I hope this has helped.

[Curious Calculus Student No.5136838]

>>5136661
>>5136543
>>5136578
>>5136549
>>5136516
Thank you guys for all the help.
Sorry I wasn't available for a moment, I took the time to eat dinner.

I will watch all derivatives-related KA videos and then look at >>5136661. Next I will read
>http://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/implicitdiffdirectory/ImplicitDiff.html
and listen to >>5136578. I will end it all by attempting all of my questions and then the problem set.

Hopefully I will get this all done by tonight.

Once again, thanks and take care.


(I will keep this thread up because of the fact that many HS seniors/college freshmen may want to ask basic Calculus questions like mine.

Have a wonderful night, everyone.
Ciao.