Use logarithmic differentiation to find $\frac{dy}{dx}$ of $y=(\ln x)^{\ln x}$ I've been asked to find the derivative of $\ln x^{\ln x}$ using a method called "logarithmic differentiation". I'm not familiar with this method of differentiation, and a search on Wikipedia doesn't tell me anything I can understand. I know that $\frac d{dx}\ln x=\frac1x$, the Inverse Function Theorem $\left(\frac d{dx}f^{-1}{'}=\frac1{f'(f^{-1}(x))}\right)$ and the Chain, Addition, Subtraction, Product and Quotient Rules. Can anybody help me with this question?
 A: If $y = \ln(x)^{\ln(x)}$, then $\ln(y) = \ln\left(\ln(x)^{\ln(x)}\right) = \ln(x) \cdot \ln(\ln(x))$. The right hand side is straightforward to differentiate, using chain and product rules. I'll leave that to you.
The left side can be differentiated with the chain rule, remembering that $y$ is a function of $x$. We do the usual: differentiate the outer function $\ln$, leaving the inner function $y$ in tact, then multiply by the derivative of the inner function. That is,
$$\frac{\mathrm{d}}{\mathrm{d}x} \ln(y) = \frac{1}{y} \cdot y' = \frac{y'}{\ln(x)^{\ln(x)}}$$
So, differentiating both sides yields
$$\frac{y'}{\ln(x)^{\ln(x)}} = \frac{\mathrm{d}}{\mathrm{d}x} (\ln(x) \cdot \ln(\ln(x)),$$
hence
$$y' = \ln(x)^{\ln(x)} \cdot \frac{\mathrm{d}}{\mathrm{d}x} (\ln(x) \cdot \ln(\ln(x)).$$
A: $$y=(\ln(x))^{\ln(x)}$$
$$\Longrightarrow \ln(y)=\ln(x)\ln(\ln(x))$$
$$\Longrightarrow \frac{y'}{y}=\frac{\ln(x)}{x\ln(x)}+\frac{\ln(\ln(x))}{x}$$
$$\Longrightarrow y' = \frac{(\ln(x))^{\ln(x)}}{x}(1+\ln(\ln(x)))$$
which is the required answer.
A: I think that this question is solved by this method : $u(x)^{v(x)}= u(x)^{v(x)}. [ln(u(x)).v(x)]'$ . Then,
$ln(x)^{ln(x)}.[ln(lnx).ln(x)]'$.
