/sci/ - Science & Math

first pull out any constants like that bottom 2
second make the negative powers positive
expand the powers and cancel

cheers for that

thank you

>>2905014


does it leave you with 4a^5

?

I think it is 4^1

bump please

(2ab)^-2 / (2(ab)^-3)
=(2^-2 / 2) * (ab)^-2 / (ab^-3)
= 1/8 * (ab)^(-2 - (-3))
= 1/8 * ab = ab/8

>>2905104


nah its

(2ab)^-2 / 2(ab)^-3

>>2905118
same thing i started with.. unless you mean

((2ab)^-2 / 2) * (ab)^-3
?

need to use brackets properly or the question is ambiguous

(2ab)^-2 / 2(ab)^-3

= 1/2 * (2ab)^-2 / (ab)^-3
= 1/2 * (ab)^3 / (2ab)^2
= 1/2 * a^3b^3 / 4a^2b^2
= 1/2 * ab / 4
= 1(ab) / 2(4)
= ab/8

<span class="math">\frac{(2ab)^{-2}}{2(ab)^{-3}}[/spoiler]
<span class="math">\frac{1}{2}(\frac{a^3b^3}{4a^2b^2})[/spoiler]
<span class="math">\frac{1}{8}(ab)[/spoiler]

Question for you math junkies: is what I did legit? Switch the numerators and denominators?

More to the point: if I have something like this

<span class="math">\frac{1}{2^{−1}}[/spoiler]

Is that the same as 2? I assume it is.

More to the point...

Chain rule on trig functions: how the fuck does it work?

>>2905160
That's wrong.

The bottom line is 2(ab)^3.
That means, you cube the a and the b, but you leave the 2 alone. The 2 is a constant, and can be factored out as 1/2. Then, subtract like powers. Then, multiply the end result by 1/2 and you get ab/8.

>>2905136

there are no brackets on the dom apart from the (ab)

(2ab)^-2
------------
2(ab)^-3
>>2905144

how come 1/2

my brain is not working this afternoon

lol sigh

(1/8(ab)^3)/(2/(ab)^2)

1/16(ab)

>>2905178
>>2905174

op here no I get it know

thank you guys very much

ab/8

>>2905178
Yes. It is legit.

1/2^-1 = 2^1/1 = 2/1 = 2

So yes.
>>2905184
Because having a 2 on the bottom of a fraction there is like 1/2. Write it out.

1/2 * a/b. What's that equal? 1(a) / 2(b) = a/2b. Right?
So since there's a 2 on the denominator there not touched by the ^3, you can factor it out as 1/2. Try it without factoring out the 2.

>>2905207

Thanks. Also, I'm going to just ask a HS physics student: can you explain the chain rule for trig functions?

You know how the normal chain rule would be...

<span class="math">(x-1)^{-2}[/spoiler]
<span class="math">(-2)*(x-1)^{-3}*(1)[/spoiler]

Right? So you take the exponent of the quotient and bring it to the front, subtract one from the quotient's exponent, and then multiply the entire thing by the derivative of the inside. I get that.

But how do you translate that into trig functionese?

<span class="math">\sin^3(x)[/spoiler]

<span class="math">3*\sin^2(x)*\cos(x)[/spoiler]

Easy so far. But then you get something like:

<span class="math">\int\sin^5(2x)\cos(2x)dx[/spoiler]

Yes I understand there is a specific rule for even and odd powers of cosine and sine. But those 'rules' don't make much sense to me. What is happening to the trig functions?

>>2905274

K how about a non-HS physic student tackle this explanation.

>>2905274

S(sin^5(2x)cos(2x))dx

Let u = sin(2x):
du = 2cos(2x)dx

S(sin^5(2x)cos(2x))dx = (1/2)S(u^5)du
= u^6/12
= sin^6(2x)/12

I wasn't really asking for the answer. Let's try something different:

<span class="math">\sin^3((2x)^2))[/spoiler]

Basically you have to keep going deeper and deeper...confuses me. What is the end goal for all of this? What are we trying to do?

<span class="math">\sin^3(4x^2)[/spoiler]

<span class="math">\sin^2(4x^2)*8x[/spoiler]

<span class="math">8x\sin^2(4x^2)[/spoiler]
<span class="math">u = \sin(4x^2)[/spoiler]
<span class="math">u' = \sin(4x^2)*8x[/spoiler]

You can see where I'm getting stuck. Any help?

>>2905786

See I think the problem is defining the 'derivative of the inside' for trig functions. It's not just a variable.

<span class="math">3*\sin^2(4x^2)*\frac{d}{dx}\sin(4x^2)[/spoiler]

<span class="math">\sin(4x^2)[/spoiler]
<span class="math">u' = \cos (4x^2)*8x[/spoiler]

Right here. Feels like I'm doing something wrong by not substituting u back into the original equation. Would this really be a u-sub step?

<span class="math">3\sin^2(4x^2)*\cos(4x^2)*8x[/spoiler]
<span class="math">24x\sin^2(4x^2)*\cos(4x^2)[/spoiler]

It's that last step...