Differentiate $\:\textrm{ln}\sqrt{\textrm{ln}\:x}$ I am confused about the solution and method of differentiating this function:
$$\frac{d}{dx}\:\textrm{ln}\sqrt{\textrm{ln}\:x}$$
Why is ln not considered a constant and then multiplied by the derivative of$\:\sqrt{\textrm{ln}\:x}$ ?
The solution is given as: 
$$\left(\frac{1}{2x\:\textrm{ln}x}\right)$$
How exactly is the chain rule applied to the entire function at once?
 A: You are misreading.
You have misinterpreted the expression as being the same thing as
$$ \ln \times \sqrt{\ln x} $$
which involves a multiplication, but that's totally wrong. It means
$$ \ln\left( \sqrt{\ln x} \right) $$
that is, the function $\ln$ is being evaluated at the value $\sqrt{\ln x}$.
A: What you have here is a function in a function.  In particular, we write
$$
f(x) = \ln(\sqrt{\ln(x)}) = \ln([\ln(x)]^{1/2})
$$
In fact, it would be easier to go one step further using the rules of logarithms and rewrite this as $f(x) = \frac 12 \ln(\ln(x))$, but I'll leave this step out so that we can emphasize the chain rule.
We see that $\ln(\sqrt{\ln(x)})$ can be broken into a composition of two functions.  In particular, we're applying $ln(\cdot)$ to $\sqrt{\ln(x)}$.  The chain rule tells us that since the derivaitive of the outer function is given by $ [\ln(x)]' = 1/x$ we can write
$$
f'(x) = \frac{1}{\sqrt{\ln(x)}} \cdot \left[ \sqrt{\ln(x)}\right]'
$$
Now, we need to find the derivative $\left[ \sqrt{\ln(x)}\right]'$.  We see that this can also be broken down into the composition of functions.  In particular, this is $\sqrt{\cdot}$ applied to $\ln(x)$.  The derivative of the outer function is given by $[x^{1/2}]' = \frac 12 x^{-1/2} = \frac{1}{2\sqrt{x}}$.  So, the chain rule tells us we can rewrite
$$
\left[ \sqrt{\ln(x)}\right]' = \frac{1}{2\sqrt{\ln(x)}}\cdot [\ln(x)]'
$$
All together, we have
$$
f'(x) = \frac{1}{\sqrt{\ln(x)}} \cdot \left[ \sqrt{\ln(x)}\right]' = 
\frac{1}{\sqrt{\ln(x)}} \cdot \frac{1}{2\sqrt{\ln(x)}}\cdot [\ln(x)]' \\
=\frac{1}{\sqrt{\ln(x)}} \cdot \frac{1}{2\sqrt{\ln(x)}}\cdot \frac 1x =
\frac{1}{2x \ln(x)}
$$
A: Define the new variables
$$u = \ln x $$
$$v = \sqrt u$$
$$w = \ln v$$
To see why you'd do this, try expressing $w$ in a different way using the previous two identities.
By applying the iterated chain rule, your derivative expression becomes
$$\frac{dw}{dx} = \frac{dw}{dv} \frac{dv}{du} \frac{du}{dv}$$
Be careful not to confuse function composition, which is all the chain rule's about, with multiplication by a constant.
A: $$y=\ln(\sqrt{\ln x})$$
$$e^{2y}=\ln x$$
$$e^{2y}(2y')=\frac{1}{x}$$
$$y'=\frac{1}{2xe^{2y}}=\frac{1}{2x\ln x}$$
