Calculus Help

[Liokae]

One problem on my homework that's bugging the hell out of me. It's not a major worry because homework is only 5% of our average and we go over it before turning it in anyway, but the problem itself is annoying me because I can't even figure out to start. This would be:

Differentiate:

x^(cosh(x))

[kevinok]

Do you mean x to the cosin times x???

[Liokae]

No, it would be x to the hyberbolic cosine of x. I got a working method off the GameFAQs boards...

Basically, thus:

y = x^cosh(x)

ln(y) = ln[x^cosh(x)]

ln(y) = cosh(x)ln(x)

y = e^[cosh(x)ln(x)]

dy/dx = e^[cosh(x)ln(x)]*[cosh(x)/x + ln(x)sinh(x)]

[Afterburner]

no he means hyperbolic cosine

i will get back to you tomorrow

[kevinok]

I know he just said that.

[Liokae]

You post four minutes after me and miss the fact that I already got the method to solve it?

[kevinok]

Okay well sorry I am no help.

[Afterburner]

Yeah, sorry, I opened the window then had to do something else for a couple minutes.

Oh well.

[Diamond187]

That's a really interesting way, cuz I know that ln(x) trick, but I've always differentiated it while they were still both in ln(x) form so I'd get a bizarre y term in there as well. I'm glad you posted that, it'll make my answers a whole lot clearer and less stupid.

[Liokae]

Very useful to remember, yes. Remember, a log is the inverse of an exponential. Natural log, which is log base e, works like this: For the ln x = u, e^u = x. for log x = u, it would be 10^u = x.

[kevinok]

And e is like pi it never ends.

[Diamond187]

there is actually a way to calculate it (besides pushing e^1 on ur calculator) but I can't remember the formula. I know it's a limit as x goes to infinite of some rational expression but I can never remember it, does anybody know it?

[Afterburner]

the limit as x goes to infinity of (1 + 1/x)^x

[Diamond187]

Yes, that's it, thnx.