Differentiate x^x - Get Help Here!

[pavadrin]

Hey,
Is it possible to differentiate x^x? I’ve tried with the use of logs, but it doesn’t seem to work…….any help would be kindly appreciated
Thanks,
Pavadrin

 

[star.torturer]

you ahve to say what yiou ahve done first ie workings and the such like

 

[fargoth]

f(x)=x^x=e^{xln(x)}

when you differentiate e, it stays the same, but you multiply by the inner derrivative:

\frac{df(x)}{dx}=(ln(x)+1)e^{xln(x)}=(ln(x)+1)x^x

 

[HallsofIvy]

First why is a question about the derivative in "Precalculus"? It sounds pretty "calculus" to me!

Logs is, of course, the way to go. If y= x^x, then log y= x log x. Differentiate both sides:
\frac{d log y}{dx}= \frac{1}{y}\frac{dy}{dx}
of course. And
\frac{d(x log x)}{dx}= log x+ \frac{x}{x}= log x+ 1

Here's an interesting point: to find the derivative of f(x)^{g(x)} there are two obvious mistakes you could make:

1) Treat the exponent, g(x), as if it were a constant and write
\frac{f(x)^{g(x)}}{dx}= g(x)f(x)^{g(x)-1}\frac{df}{dx}

2) Treat the base, f(x), as if it were a constant and write
log(f(x))f(x)^{g(x)}\frac{dg}{dx}

Because those are mistakes, neither is, of course, correct. The correct derivative of f(x)^{g(x)} is the sum of those!

 

[fargoth]

he did say, he tried logs...

 

[Gelsamel Epsilon]

Just because he tried logs doesn't mean he did it right. HallsofIvy way is the correct way to do it. Log both sides then differentiate both.

 

[fargoth]

yeah, you're right, we should have lead him to the solution and not just give it out like that...

anyway, doing it my way or differentiationg both sides is really the same thing... i didn't need to look at both sides because i just used the equation a^b=e^{bln(a)}
so y is still y... so if i differentiate both sides i get:
dy=(ln(x)+1)e^{xln(x)}dx

 

[power freak]

Instead of starting a new thread I thought I'd "borrow" this one:

I am trying to differentiate the function y=(x^x)^x

Would this working be correct?

y = (x^x)^x = x^f^(^x^)

\rightarrow ln(y) = f(x)ln(x)

\frac{dy}{dx}\frac{1}{y} = (f(x))(\frac{dln(x)}{dx}) + (ln(x))(\frac{df(x)}{dy})

= \frac{f(x)}{x} + f'(x) lnx

\rightarrow \frac{dy}{dx} = y((\frac{f(x)}{x}) + f'(x)ln(x))

But since:

f(x) = x^x

f&#039;(x) = x^x(ln(x) + 1)<br />

(I've spared the working for that since it's covered up there)

So does that make:

\frac {d}{dx} (x^x)^x = (x^x)^x ( \frac{x^x}{x} + (ln(x)+1)x^x)<br />

??

Or and I missing something along the way? This is my first time using the product rule so I want to make sure I'm doing it correctly..

Thanks,

Lewis

 

[VietDao29]

Instead of starting a new thread I thought I'd "borrow" this one:

I am trying to differentiate the function y=(x^x)^x

Would this working be correct?

y = (x^x)^x = x^f^(^x^)

Nope, this line is wrong. Instead, it should read:
y = {(x ^ x)} ^ x = f(x) ^ x, where f(x) = x^x. :)
Ok, let's take log of both sides:
\ln y = x \ln f(x)
Differentiate both sides with respect to x, we have:
\frac{y&#039;_x}{y} = \ln f(x) + x \frac{f&#039;(x)}{f(x)} \Rightarrow y&#039;_x = y \left( \ln f(x) + x \frac{f&#039;(x)}{f(x)} \right) = {(x ^ x)} ^ x \left( \ln f(x) + x \frac{f&#039;(x)}{f(x)} \right), right?
And from the above posts, we have:
f'(x) = (x^x)' = (x^x) (ln(x) + 1). So plug that into the expression above, we have:
y&#039;_x = {(x ^ x)} ^ x \left( \ln f(x) + x \frac{x ^ x (\ln (x) + 1)}{x ^ x} \right) = {(x ^ x)} ^ x \left( \ln (x ^ x) + x (\ln (x) + 1)} \right)
= {(x ^ x)} ^ x \left( \ln (x ^ x) + \ln (x ^ x) + x)} \right) = {(x ^ x)} ^ x \left( 2 \ln (x ^ x) + x)}
Ok. Can you get this? :)

 

[StatusX]

Lewis, your way would be correct if you wanted to differentiate y=x^{(x^x)}. But as you have it written now, y=(x^x)^x=x^{(x^2)}.

 

[power freak]

Lewis, your way would be correct if you wanted to differentiate y=x^{(x^x)}. But as you have it written now, y=(x^x)^x=x^{(x^2)}.

That's what I was trying to write! I can't get the hang of tex at all... I editted it a few times but couldn't get it to look like: y=x^{(x^x)}

Thanks though at least I know I'm on the right path (I should have explained what I was trying to do a little better in hindsight..)

I'll try and tackle the:

y=(x^x)^x=x^{(x^2)}

Problem now and see if I can get Vietdao's solution. :)


Lewis