/sci/ - Science & Math

sin^2(2x)dx

I support this.
<span class="math">f(x) = x^3 + 2x + 1[/spoiler]

>>1764564
How do this?
All i learned was f(x)=h(g(x) problems, could you give me some of those? I would like to solve yours if i knew how :x
>>1764565
computing

>>1764565
f1(x)= (3x +2 )

>>1764585
wrong

>>1764585

Wrong it's 3x^2+2

Holy shit how can one fail at the power rule?

>>1764593
whoops! i forgot to write that, my bad, yeah i had the right answer

f1(x) = 3x^2 +2

>>1764591
>>1764593

He clearly said that he just learned derivation and needs practice, dickfucks.

>>1764599
it's okay lol, i expect the trolls

>>1764598
nice work. Try this one:

<span class="math">f(x) = (x^2 + 2)^2[/spoiler]

maybe some more like these ones:

f(x) =h1(x) * g1(x) * h(x) * g1(x)

>>1764612
Why don't you just write ` instead of 1

>>1764564
Trying this in my head here, almost certainly will be wrong.
4sin(2x)cos(2x)?

>>1764605
f(x) = (x^2 +2)^2
f1(x) = 2 * 2 (x^2+2)^2
f1(x) = 4(x^2 +2)^2

i think i got that right

>>1764579
>>1764564
Just because i'm a nice guy, you can rewrite this one as
<span class="math"> \frac{1 - cos(2x)}{2} [/spoiler]
Easier for you?

>>1764617
kay i will

>>1764623
no idea what cosine is :x

Explain?

>>1764612
How about this: I will tell you that the derivative of sin(x) is cos(x), and the derivative of cos(x) is -sin(x), in case you haven't learned it yet. So then,

<span class="math">f(x) = \sin(x) \cdot \tan(x)[/spoiler]

>>1764623
It should be 4x, not 2x.

>>1764629
No offense, but you probably shouldve learned algebra II or precalculus before jumping into calculus

>>1764621
You did not, but you were going in the right direction.

>>1764629
it's a ratio in right triangles. given an angle measure, its cosine will be the ratio between the side adjacent to it and the hypoteneuse

>>1764621
My corrections


f(x) = (x^2 +2)^2
f1(x) = 2x * 2 (x^2+2)
f1(x) = 4x(x^2 +2)

>>1764630
okay so . . .
in response to: >>1764564
it'd be:
f'(x) = 2cos uhmmm
f'(x) = cosx^2 * 2 * dx?

>>1764579
>How do I do this?
If you JUST learned derivatives, this is going to give you trouble. The trick is in the "chain rule."

I asked you to find the derivative of:
[sin(2x)]^2
Well the most "outward" part of this is the "^2." So what's the derivative of N^2? It's 2N, but since N is not necessarily x, you also have to multiply the answer by the derivative of N.

So the derivative of sin^2(2x) is 2sin(2x) * sin(2x)dx.
So what's the derivative of sin(2x)dx?
The derivative of sinx is cosx. So the derivative of sinN is cosN, then times the derivative of N.

So the derivative of sin(2x) is cos(2x) * (2x)dx.
I assume you know the derivative of 2x is 2, so now we can write out the entire answer:

2sin(2x) * cos(2x) * 2 = 4sin(2x)cos(2x). You can further substitute the sin(2x) and cos(2x) terms with their identities if you want everything in terms of x instead of 2x, but the above answer is sufficient.

>>1764636
okay
>>1764637
Im in algebra II right now (as a senior), because I didn't do well in early high school math :(, learning so i can get into Honors Physics by 2nd quarter
>>1764638
i saw
>>1764640
just read that, I'll watch some khan on this tomorrow
>>1764644
thank you

>>1764646
literally just learned them 5 minute ago :P, I'm getting a calculus textbook tomorrow so maybe that'll help me

Oh /sci/ look at all together helping this budding mathemagican. It brings warmth of my heart it does.

>>1764661
haha, I'm glad I have this as resource honestly, you guys are really helpful sometimes when it comes to learning math (personally)

>>1764655
Looks like you need some work on derivatives of compositions. These are fairly common as you go on, so keep practicing. Try this:

<span class="math">f(x) = (5x^2+3x + 1)^3[/spoiler]

I have a gift for OP.

http://www.bluffton.edu/~nesterd/java/derivs.html

If anyone knows of something similar for Integrals, let me know.

>>1764619
Yes!

OP, I just browsed through the comments and noticed that you don't yet know what sine or cosine MEAN. At least one poster said that you should learn these before messing with calculus, and that's extremely true. Pre-Calculus deals a lot with those trig functions and their relationships. Physics uses them a lot. Physics uses calculus a lot. In short, you don't learn these, you're not going far.

Sine and cosine are functions of angles. Imagine a circle of radius 1 centered on the origin of an x/y plane. This is called the unit circle. We will draw angles through it. The vertexes of the angles are at the origin, and the base of any angle is along the positive x axis. The angles open counter-clockwise.

Wherever the angle lands on the circle, examine the point there. Its x coordinate is cosine, and its y coordinate is sine. Therefore, the sine of 90 degrees is 1, and the cosine of 90 degrees is 0. The cosine of 180 degrees is -1. No sine or cosine values will be above 1 or below -1.

Pic related.

>>1764666
f'(x) = (10x + 3) * 3(5x^2+3+1)^2

>>1764677
I will learn these ASAP then :O, tomorrow I mean. I have Algebra II homeowork . . .

thank you though, it's greatly appreciated

>>1764678
:) Well done!

Let's revisit the trig ones above. I gave you:
<span class="math">D_x \cos(x) = -\sin(x)[/spoiler]
<span class="math">D_x \sin(x) = \cos(x)[/spoiler]
now I will tell you
<span class="math">\tan(x) = \frac{\sin(x)}{\cos(x)}[/spoiler]
Even though these may be new, with this information you should be able to find the derivative of
<span class="math">f(x) = \sin(x) \cdot \tan(x)[/spoiler]

>>1764690
I'll tell you right now I won't be able to do that one. I have no idea how to plug these in, like:

does the D_x \cos(x) = -\sin(x) mean that sin = -dxcos(x)?

>>1764703
Not sure what you're asking. How about an example?
<span class="math">D_x \sin(3x) = 3\cos(3x)[/spoiler]

Better?

>>1764703
<span class="math"> D_x cos(x) = -sin(x) [/spoiler]
is a way of saying the derivitive of cos(x) is -sin(x).

>>1764727
still greek to me :/ I'm going to watch "introduction to limits pt 1-x" tomorrow so maybe it'll cover that. if not the textbook will.

thanks though

man, why can't you be asking for derivatives?
they're so much easier to troll with.
alright:
erf(x)
look it up.

I'm not sure if the OP has jsMath, so the question you were giving him about the sines and cosines probably looked fucked up to him. I could be wrong though.

>>1764741
what is that?
Anymore practices similar to: >>1764666?

those are really helpful in practicing for me

thanks

>>1764749
if you're referring to a program I probably don't have it because I'm on mint atm, and no graphing calculator, I've been writing this all out

I am a newb and don't know how to respond to someone's post, but here is an online integral solver that someone requested


http://integrals.wolfram(.)com/index.jsp

>>1764752
http://en.wikipedia.org/wiki/Error_function

>>1764752
Sure.

<span class="math">f(x) = (x^3 + (2x+1)^2 + 1)^2[/spoiler]

<span class="math">f(x) = \frac{x+1}{x-1}[/spoiler]

<span class="math">f(x) = (\frac{x+1}{x-1})^2[/spoiler]

>>1764682
Good. It'll take a bit to study, and pre-calc is indispensable. But, know your trig functions, their derivatives, the inverse trigs, THEIR derivatives, and the identities.

Pic helpful: it's the plot of all the values of sine and of cosine when you round the unit circle once.

f(x)=(x3+(2x+1)2+1)2
i think, but I'm not sure:
f'(x) = 2 * (3x^2) * 2(x^3(2x+1)+1)^2
i can't remember i think i do inner, then x^2, then the )^2 right?

>>1764823
WAIT I FOUND MY EXAMPLE HANG ON

>>1764802
You forgot hyperbolic trig functions. You need their derivatives too.

>>1764788
okay let's see

f'(x) = 4 (2x+1) + (3x^2) *2(x^3(2x+1)^2))^2

the +1 after ")" through me off a bit but i think it just becomes 0 right?

>>1764823
There's actually a few levels to this one. First you have the outside to deal with, so (D_x means "derivative of", ok?)
<span class="math">D_x f(x) = 2(\ldots) \cdot ?[/spoiler]
The '?' will be the derivative of whatever's inside.

What's inside is
<span class="math">g(x) = x^3 + (2x+1)^2 + 1[/spoiler]
That's not too bad. It should be (the D_x means "derivative of" remember!)
<span class="math">D_x g(x) = 3x^2 + 2\cdot (2x+1)\cdot 2 = 3x^2 + 4(2x+1) = 3x^2 + 8x + 4[/spoiler] Now we can take that back to the original.
D_x f(x) = 2(x^3 + (2x+1)^2 + 1)(3x^2 + 8x + 4) which may or may not be simplified more to your taste.

>>1764857
I think i did it wrong . . . I did the g'(x) part first to get the inner (2x+1) = 2x part right?

>>1764863
Yes, you are mixing up what's called the chain rule. There's a couple different ways to picture it. I will write two, maybe one will make it click
<span class="math">D_x f(g(x)) = f\backprime (g(x)) \cdot g\backprime (x)[/spoiler]
<span class="math">\frac{d}{dx}f(g(x)) = \frac{df}{dg} \frac{dg}{dx}[/spoiler]

So it is like we create the temporary variable.
If
<span class="math">f(x) = (5x+5)^2[/spoiler] then we'll call g(x)
<span class="math">g(x) = 5x+5[/spoiler]
With that, we can say
<span class="math">f(g) = g^2[/spoiler] and
<span class="math">D_g f(g) = 2g \cdot D_x g[/spoiler]
Then <span class="math">D_x f(g(x)) = 2(5x+5) \cdot 5x[/spoiler]

This chain rule is extremely important. 99% of basic calc can be forgotten and rederived if you can master the chain rule.

>>1764908
correcting this, then I will delete it

>>1764863
Yes, you are mixing up what's called the chain rule. There's a couple different ways to picture it. I will write two, maybe one will make it click
D_x f(g(x)) = f'(g(x)) g'(x)
<span class="math">\frac{d}{dx}f(g(x)) = \frac{df}{dg} \frac{dg}{dx}[/spoiler]

So it is like we create the temporary variable and say: "The derivative of f with respect to x is the derivative of f with respect to g times the derivative of g with respect to x."
If
<span class="math">f(x) = (5x+5)^2[/spoiler] then we'll call g(x)
<span class="math">g(x) = 5x+5[/spoiler]
With that, we can say
<span class="math">f(g) = g^2[/spoiler] and
<span class="math">D_g f(g) = 2g \cdot D_x g[/spoiler]: The derivative of f with respect to g...
<span class="math">D_x g(x) = 5[/spoiler]: The derivative of g with respect to x...
Then <span class="math">D_x f(g(x)) = 2(5x+5) \cdot 5x[/spoiler]: the derivative of f of g of x is the derivative of f with respect to g times the derivative of g with respect to x.

This chain rule is extremely important. 99% of basic calc can be forgotten and rederived if you can master the chain rule.

>>1764908
correcting this, then I will delete it

>>1764863
Yes, you are mixing up what's called the chain rule. There's a couple different ways to picture it. I will write two, maybe one will make it click
D_x f(g(x)) = f'(g(x)) g'(x)
<span class="math">\frac{d}{dx}f(g(x)) = \frac{df}{dg} \frac{dg}{dx}[/spoiler]

So it is like we create the temporary variable and say: "The derivative of f with respect to x is the derivative of f with respect to g times the derivative of g with respect to x."
If
<span class="math">f(x) = (5x+5)^2[/spoiler] then we'll call g(x)
<span class="math">g(x) = 5x+5[/spoiler]
With that, we can say
<span class="math">f(g) = g^2[/spoiler] and
<span class="math">D_g f(g) = 2g \cdot D_x g[/spoiler]: The derivative of f with respect to g...
<span class="math">D_x g(x) = 5[/spoiler]: The derivative of g with respect to x...
Then <span class="math">D_x f(g(x)) = 2(5x+5) \cdot 5[/spoiler]: the derivative of f of g of x is the derivative of f with respect to g times the derivative of g with respect to x.

This chain rule is extremely important. 99% of basic calc can be forgotten and rederived if you can master the chain rule.

>>1764828
Nope, didn't forget them--just never heard about them :-O
Looking them up now...

>>1764944
yeah, i knew something was f-ed up in my chain >.< i do f(x) = h(g(x)) by the way not d/d

im going to bed but thank you everyone for the practice and intellectual stimulation ^^