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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?

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    $\begingroup$ $\ln$ is not a constant, it is a function. Saying that $\ln$ is a constant would mean saying, for example, that $$ \frac{d}{dx} \ln (x) = \ln \frac{d}{dx} x = \ln $$ $\endgroup$ – Omnomnomnom Nov 6 '16 at 1:16
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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 $\sqrt{\ln x}$.

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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 $$ f'(x) = \frac{1}{\sqrt{\ln(x)}} \cdot \left[ \sqrt{\ln(x)}\right]' = \frac{1}{\sqrt{\ln(x)}} \cdot \frac{1}{2\sqrt{\ln(x)}}\cdot [\ln(x)]' \\ =\frac{1}{\sqrt{\ln(x)}} \cdot \frac{1}{2\sqrt{\ln(x)}}\cdot \frac 1x = \frac{1}{2x \ln(x)} $$

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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.

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$$y=\ln(\sqrt{\ln x})$$ $$e^{2y}=\ln x$$ $$e^{2y}(2y')=\frac{1}{x}$$ $$y'=\frac{1}{2xe^{2y}}=\frac{1}{2x\ln x}$$

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