Can someone help me to derivate x^x^x? I'm starting to belive it's not possible... : D heeelp! Thank you in advance!
Oh, it's perfectly possible, and fun too .
I have class soon, but I will get you started.
First, what is the derivative of x^x?
We have y = x^x
Note: x > 0
ln(y) = ln(x^x)
ln(y) = x*(ln(x))
Differentiate both sides with respect to x;
y'(1/y) = ln(x) + x*(1/x) = (ln(x) + 1)
y' = (ln(x) + 1)*y
But recall y = x^x
y' = (ln(x) + 1)*x^x
Now you can use this to find the derivative of x^x^x.
Hello, kukkeball!
Here's another approach . . .
Differentiate: . $\displaystyle y \:=\:x^{x^x}$
Take logs: .$\displaystyle \ln y \;=\;\ln\left(x^{x^x}\right) \;=\;x^x\ln x$
Take logs again: .$\displaystyle \ln(\ln y)\;=\;\ln\left[x^x\ln x\right] \;=\;\ln\left(x^x\right) + \ln(\ln x) \;=\;x\ln x + \ln(\ln x)$
Differentiate implicitly (very carefully) . . .
. . $\displaystyle \frac{1}{\ln y}\cdot\frac{1}{y}\cdot\frac{dy}{dx}\;=\;x\cdot\fr ac{1}{x} + \ln x + \frac{1}{\ln x}\cdot\frac{1}{x}$
We have: .$\displaystyle \frac{1}{y\ln y}\cdot\frac{dy}{dx} \;=\;1 + \ln x + \frac{1}{x\ln x}$
Then: .$\displaystyle \frac{dy}{dx}\;=\;y\ln y\left(1 + \ln x + \frac{1}{x\ln x}\right)$
Therefore: .$\displaystyle \frac{dy}{dx}\;=\;x^{x^x}\ln\left(x^{x^x}\right)\l eft(1 + \ln x + \frac{1}{x\ln x}\right)
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