x^x

It looks to me like you would get it from u-sub.

I'm sure someone on google has gone through the steps though.

File: 288px-Lambert-w.svg.png (7KB, 288x229px) Image search: [Google]

7KB, 288x229px

>>7982335
This """"""""Function"""""""" may help you

>>7982335
I didn't get to logarithm's and exponent's in Calculus that's my next unit after tomorrows test.
Can you please show me why y = sin(kx) has the derivative of y'= k*cos(kx)

>>7982347
I know it has something to do with chain rule but I don't know how to apply it to this situation.

x^x=e^(x*lnx)
Derivative of e^(x*lnx)....
(lnx+x/x)*(e^(x*lnx))
(Lnx+1)*x^x

>>7982351
so obvious once I see it, thx Anon

>>7982347
Because that's the chain rule.

You have an outer function (sine) and an inner function (kx). Therefore the derivative is the derivative of the outside (derivative of sine is cosine), leave the inside alone, multiply by the derivative of the inside (derivative of kx is k).

>>7982347
This is the chain rule. You should fucking know this by now.

>>7982335
Take the log of both sides.
Then you get ln(f(x)) = ln(x^x), which implies that ln(f(x)) = xln(x) by a basic property of logarithms (you can google this). Take the derivative of both sides. You should know that the derivative of ln(x) = 1/x (again, you can google this.) Thus, you apply the chain rule (google this if you don't know it) to get that f '(x)/f(x) = ln(x) + 1. This implies, by multiplying both sides of the equation by f(x), that f '(x) = f(x)(ln(x) + 1). But recall that f(x) = x^x as this was the problem statement.
Thus f '(x) = x^x (ln(x) + 1) and we have demonstrated it.

Fuck I miss the intuitive calculus days.

Look up its power series.

>>7982356
But you can't log a negative number, so isn't some information lost or some thing?

>>7982354
Thanks, I understand it now.

>>7982358
Nop

>>7982347
chain rule: let u=kx, so sin(kx) = sin(u) and
[math] \displaystyle \frac{ \mathrm{d}}{ \mathrm{d}x} \sin u=( \cos u) \left ( \frac{\mathrm{du} }{ \mathrm{d} x} \right)[/math]

>>7982343
Actually, it won't. I don't get why you insist on posting that """"function"""" crap all the time.

>>7982335
[math]\displaystyle x^x = e^{x \ln(x) } \\ g(x)=x, h(x)=\ln(x), g'(x)=1, h'(x)=x^{-1} \\ u=g(x)h(x), u'=g'(x)h(x)+g(x)h'(x)=\ln(x)+1 \\ \frac {\mathbb{d}} {\mathbb{dx}}e^u = e^u * u', e^u=x^x, u'=\ln(x)+1, \therefore \frac {\mathbb{d}} {\mathbb{dx}}e^u = x^x (\ln(x)+1)[/math]

>>7982356
niice

>>7982371
good work on the LaTeX, thx Anon

File: log(z).png (2KB, 554x25px) Image search: [Google]

2KB, 554x25px

>>7982383
Everything looks better in [math]\LaTeX[/math]

>>7982360
+1 for remembering that property. This is, however, somewhat incorrect. Logs are defined for negative numbers, though. I'll show you why.

e^x has a Taylor series equal to [math]\sum_{i=0} \frac {x^n} {i!} = \frac {x^n} {1} + \frac {x^n} {1} + \frac {x^n} {2} + \frac {x^n} {6} + \frac {x^n} {24} + \frac {x^n} {120}...[/math]
If you want to see why, just look it up, there's lots of good proofs of it.

The unique property of this series is that we can plug complex numbers into it. By setting x=i*b, we get the series [math]e^{bi} = 1+bi-\frac {b^2} {2} + \frac {ib^3} {6} ...[/math]

Because this series converges absolutely, we're allowed to reorder this. I don't want to make two separate posts so I'm going to skip how one part is i*sin(x) and one is cos(x), and jump right to the conclusion: [math]e^{ix}=\cos(x)+i\sin(x) [/math] Cosine and sine have a range of [-1,1], so e^ix can only go from -1 to 1. It may not be immediately obvious, but e^ix represents the unit circle. A complex number [math]z:=re^{i\theta}[/math], so pic related is my next statement (stolen from https://en.wikipedia.org/wiki/Complex_logarithm ) Looking back this was a shitty post and I want to erase it and start over but I spent like 20 minutes typing this and I really don't want to restart.

tl;dr you actually can

x^x = e^[ln(x^x)] =e^(xlnx)

d/dx[e^(xlnx)]

= [d/dx(xlnx)] * [e^(xlnx)]

= [1*lnx + (1/x)*x ] * [(e^lnx)^x]

= [lnx + 1] * x^x

Fin


Crucial thing here: remember that any number can be represented as e^ln(number) and go from there.

>>7982430
I remembered that part, but was stymied thereafter, maybe fatigue