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
[tex]f(x)=x^x=e^{xln(x)}[/tex] when you differentiate e, it stays the same, but you multiply by the inner derrivative: [tex]\frac{df(x)}{dx}=(ln(x)+1)e^{xln(x)}=(ln(x)+1)x^x[/tex]
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: [tex]\frac{d log y}{dx}= \frac{1}{y}\frac{dy}{dx}[/tex] of course. And [tex]\frac{d(x log x)}{dx}= log x+ \frac{x}{x}= log x+ 1[/tex] Here's an interesting point: to find the derivative of [itex]f(x)^{g(x)}[/itex] there are two obvious mistakes you could make: 1) Treat the exponent, g(x), as if it were a constant and write [tex]\frac{f(x)^{g(x)}}{dx}= g(x)f(x)^{g(x)-1}\frac{df}{dx}[/tex] 2) Treat the base, f(x), as if it were a constant and write [tex]log(f(x))f(x)^{g(x)}\frac{dg}{dx}[/tex] Because those are mistakes, neither is, of course, correct. The correct derivative of [itex]f(x)^{g(x)}[/itex] is the sum of those!
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.
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 [tex]a^b=e^{bln(a)}[/tex] so y is still y... so if i differentiate both sides i get: [tex]dy=(ln(x)+1)e^{xln(x)}dx[/tex]
Instead of starting a new thread I thought I'd "borrow" this one: I am trying to differentiate the function [tex]y=(x^x)^x[/tex] Would this working be correct? [tex]y = (x^x)^x = x^f^(^x^) [/tex] [tex] \rightarrow ln(y) = f(x)ln(x) [/tex] [tex]\frac{dy}{dx}\frac{1}{y} = (f(x))(\frac{dln(x)}{dx}) + (ln(x))(\frac{df(x)}{dy})[/tex] [tex]= \frac{f(x)}{x} + f'(x) lnx[/tex] [tex]\rightarrow \frac{dy}{dx} = y((\frac{f(x)}{x}) + f'(x)ln(x))[/tex] But since: [tex] f(x) = x^x[/tex] [tex]f'(x) = x^x(ln(x) + 1) [/tex] (I've spared the working for that since it's covered up there) So does that make: [tex] \frac {d}{dx} (x^x)^x = (x^x)^x ( \frac{x^x}{x} + (ln(x)+1)x^x) [/tex] ?? 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
Nope, this line is wrong. Instead, it should read: [tex]y = {(x ^ x)} ^ x = f(x) ^ x[/tex], where f(x) = x^{x}. :) Ok, let's take log of both sides: [tex]\ln y = x \ln f(x)[/tex] Differentiate both sides with respect to x, we have: [tex]\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)[/tex], 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: [tex]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)[/tex] [tex]= {(x ^ x)} ^ x \left( \ln (x ^ x) + \ln (x ^ x) + x)} \right) = {(x ^ x)} ^ x \left( 2 \ln (x ^ x) + x)}[/tex] Ok. Can you get this? :)
Lewis, your way would be correct if you wanted to differentiate [itex]y=x^{(x^x)}[/itex]. But as you have it written now, [itex]y=(x^x)^x=x^{(x^2)}[/itex].
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: [tex]y=x^{(x^x)}[/tex] 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: [tex]y=(x^x)^x=x^{(x^2)}[/tex] Problem now and see if I can get Vietdao's solution. :) Lewis