Calculus Help! Edited new problem

[Scrooge2]

find function g(x) with dg/dx = [Ln(x)]/x

*Edit*

Find tan (arcsin (x))

what the heck does that mean? It didn't ask for the derivative...

 

[Stojakapimp]

oh oops, we are looking for the integral

 

[Scrooge2]

yes we are

 

[Stojakapimp]

use substitution.

Make u = Ln(x), so du = 1/x. Now you can just replace the 1/x part with du

 

[GiLtY]

Originally posted by: Scrooge2
find function g(x) with dg/dx = [Ln(x)]/x

Integrate the thing, you'll have a constant term, C. Unless you know the initial conditions you won't be able to find the exact solution.

Easiest way is to use u/du substitution, let u equal to LN(x)

--GiLtY

 

[Scrooge2]

can anyone solve it out for me? I'm not sure if i got it right or not, but I got [ln^2x]/2+c

I just haven't done this in a while...

 

[GiLtY]

Scrooge: that seems right


--GiLtY

 

[dighn]

Originally posted by: Scrooge2
can anyone solve it out for me? I'm not sure if i got it right or not, but I got [ln^2x]/2+c

I just haven't done this in a while...

verifying an integral is easy, just differentiate it. looks right to me.

 

[Scrooge2]

Yeah, I think I've got it. Thanks all. Like I said I did this last year and don't remember much of it. I'm transfering to another college and these ingrates don't want to take my calc I + II credits because they gave me some dumb spiel about their college's math department is top notch. They are a state college and wont take calc credits from a private engineering university but will take calc credits from the local community college. How dumb is that?

Anyways I'm trying to take this test so I can skip at leat calc I. It's slowly starting to return to me THANKS!

 

[Scrooge2]

New problem:

Find tan(arcsin (x))

 

[Scrooge2]

wah!

 

[WhiteKnight]

Maybe that's some sort of typo? I don't understand what they are asking for either. Perhaps you are just supposed to simplify it?

 

[Syringer]

That's a real simple problem. Just draw your triangle..since it's asking for sin(x)..

Draw a right triangle, have theta be one of the angles, and since it's sin..x represents the side opposite to the angle, and since x = x/1, 1 represents the hypotenuse, and the adjacent side would then equal sqrt(1-x^2)..through Pythagorean's theorem.

Then, all you have to take is the tangent of that, which means y/x, or x / sqrt(1-x^2)

 

[Scrooge2]

i'm not sure i understand syringer...

 

[Syringer]

Okay, I drew up a picture here..

http://pichosting.pcthike.com/albun15/untitled

That's your triangle..that's what arcsin(x/1) looks like. Since it's sine, you know that the angle theta is opposite over hyptonuse, or in this case, x/1..so the side opposite to theta is x, and the hypotonuse is 1.

From there, we can derive the adjacent side as being sqrt(1-x^2) through Pythagorean's theorem.

Then finally, since we want to find tan(arcsin(x/1)), we take the tangent of the triangle..which is opposite over adjacent..or in this case, x/sqrt(1-x^2)

Hope that was clearer

 

[Syringer]

Well, do you get it?