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:

$$f(x) = x^x$$

$$f'(x) = x^x(ln(x) + 1)$$

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

So does that make:

$$\frac {d}{dx} (x^x)^x = (x^x)^x ( \frac{x^x}{x} + (ln(x)+1)x^x)$$

??

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:
$$\frac{y'_x}{y} = \ln f(x) + x \frac{f'(x)}{f(x)} \Rightarrow y'_x = y \left( \ln f(x) + x \frac{f'(x)}{f(x)} \right) = {(x ^ x)} ^ x \left( \ln f(x) + x \frac{f'(x)}{f(x)} \right)$$, 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:
$$y'_x = {(x ^ x)} ^ x \left( \ln f(x) + x \frac{x ^ x (\ln (x) + 1)}{x ^ x} \right) = {(x ^ x)} ^ x \left( \ln (x ^ x) + x (\ln (x) + 1)} \right)$$
$$= {(x ^ x)} ^ x \left( \ln (x ^ x) + \ln (x ^ x) + x)} \right) = {(x ^ x)} ^ x \left( 2 \ln (x ^ x) + x)}$$
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