Math help, should be simple.

[Kurt_]

I know a lot of you haven't done anything close to this, but I also know some of you have.

the question asks to take the derivative of a composite function with an inverse tangent inside, along with 'e'.

Let F(x) = f(arctan[(e^x)^2])

Assuming f(x) is differentiable, find F'(x).

(It might be easier if you write that out on a piece of paper to look at it.

Now, the derivative of 'f' is stated as f'(x), where x is all the guts.

Since it is a composite, and using arctan' = 1/(1+x^2):

F'(x) = f'(arctan[(e^x)^2]) [1/(1 + (e^x)^2)] [(e^x)^2] [2x]

simplified,

F'(x) = 2xe^(x^2) / (1 + (e^x^2)^2).

What troubles me is the simplest part.

e^(x^2)^2 is equal to e^(2x^2).

I say it's equal to e^[(x^2)(x^2)] of which the exponent is equal to

2x^4.

Why is this wrong. Why is x^2^2 = 2x^2

You can see the question here (#1 on test #1, the answer is e. Yeah, I'm stuck on the first question three days from the midterm!)

http://www.math.mcmaster.ca/childsa/1z04/sample2.pdf

[SonyPortableizer]

Wow, doing even that is a great feat
A while back on ABC there was a short story on online tutors for free(first 3 lessons, then you pay if you like their work), try one of them
I dont have any links

[Black Six]

Your problem is your parentheses. It's not arctan[(e^x)^2]. It should be (based on the PDF) arctan[e^(x^2)]. Solving like that:
(arctan[e^(x^2)])' = (1 / (1 + (e^(x^2))^2) * (e^(x^2))'
= (1 / (1 + e^(2*(x^2))) * (e^(x^2)) * (x^2)'
= (1 / (1 + e^(2*(x^2))) * (e^(x^2)) * 2x
= (e^(x^2) * 2x) / (1 + e^(2*(x^2)))

In answer to your particular question:
e^(x^2))^2 = (e^(x^2)) * (e^(x^2)) = e^((x^2) + (x^2)) = e^(2*(x^2))

Hope that helps!

[Aguiluz]

I clicked it because it said "it should be simple" but I was like, WTF? when I saw the thing.

Any background about this subject and where is it used for?

[Rekarp]

I clicked it because it said "it should be simple" but I was like, WTF? when I saw the thing.

Any background about this subject and where is it used for?

engineering. We use something similar for Second order RLC circuits and the like.

[Kurt_]

BlackSix: Perfect!

[Black Six]

Of course it's perfect.

Glad I could help!