Question

Let
\[y = x^{x^x}\]
Find dy/dx.

Answer 1

dy/dx, you mean ?

Answer 2

Logarithmic differentiation.
\[y=x^{x^x}~\Rightarrow~\ln y=x^x\ln x\]
Repeat the process, then use implicit differentiation.

Answer 3

\[\begin{align*}\ln y=x^x \ln x~\Rightarrow~&\ln(\ln y)=\ln(x^x \ln x)\\
&\ln(\ln y)=\ln(x^x)+\ln(\ln x)\\
&\ln(\ln y)=x\ln x+\ln(\ln x)
\end{align*}\]
@Loser66, does that answer your question? I'm not sure what you mean by "as usual."

Answer 4

What do you do then....

Answer 5

@Jhannybean, Implicit differentiation:
\[\frac{d}{dx} \ln(\ln y)=\frac{d}{dx}\bigg[x\ln x+\ln(\ln x)\bigg]\\
\frac{\frac{y'}{y}}{\ln y}=\left(\frac{d}{dx}x\right)\cdot\ln x+x\cdot\left(\frac{d}{dx}\ln x\right)+\frac{d}{dx}\ln(\ln x)\\
\frac{y'}{y\ln y}=\ln x +1+\frac{\frac{1}{x}}{\ln x}\\
\frac{y'}{y\ln y}=\ln x +1+\frac{1}{x\ln x}\\
\frac{y'}{y\ln y}=\ln x +1+\frac{1}{x\ln x}\\
y'=y\ln y~\bigg(\ln x +1+\frac{1}{x\ln x}\bigg)\]
Then substitute back:
\[y'=x^{x^x}\ln \left(x^{x^x}\right)~\bigg(\ln x +1+\frac{1}{x\ln x}\bigg)\]

Answer 6

I may have made some mistakes in finding the derivatives throughout all that ... but this is the procedure for this type of problem.

Answer 7

No mistakes, i dont think. Just repetition of some steps but that doesnt matter

Answer 8

Holy siths, I thought the answer would be a little more ... aesthetically pleasing.
Thanks though, guys!

Answer 9

Natural logarithmic functions hardly are....