Chain rule to find derivative of $\ln(\sqrt{x})$ ok so by using the log rule where the power becomes the coefficient, we get $\frac{1}{2}\ln(x)$
$dy/dx=1/(1/2)x$
my question is,since it is a logarithmic function of a square root function of $x$, can we use the chain rule to get the same answer as above?
 A: \begin{align}
\dfrac{d}{dx}(\ln(x^{\frac{1}{2}})) & = \dfrac{d}{dx}(\frac{1}{2}\ln(x)) \\
& = \frac{1}{2} \cdot \dfrac{d}{dx}(\ln(x)) \\
& = \frac{1}{2} \cdot \frac{1}{x} \\
& = \frac{1}{2x}
\end{align}
Keep it simple, no chain rule needed here, as long as you know $\dfrac{d}{dx}(\ln (f(x))) = \frac{f'(x)}{f(x)}$.
A: Note that
$$(\ln f(x))’=\frac{f'(x)}{f(x)}=\frac{\frac{1}{2\sqrt x}}{\sqrt x}=\frac1{2x}$$
and we can use any rules for derivatives to obtain the same result, for example by product rule
$$\ln \sqrt x=\frac12\cdot\ln x=\left(\frac12\right)'\cdot \ln x+\frac12\cdot(\ln x)'=\frac1{2x}$$
or by quotient rule
$$\ln \sqrt x=\frac{\ln x}2=\frac{\frac1x\cdot 2-(2)'\cdot \ln x}{4}=\frac1{2x}$$
and so on. 
In this case the interpretation of $\ln \sqrt x=\frac12 \log x$ as 
$$(cf(x))'=cf'(x)$$
is of course the most effective and clever.
A: We can compute the derivative using the chain rule: the derivative of $\ln(x)$ is equal to $\frac{1}{x}$, the derivative of $\sqrt{x}$ is $\frac{1}{2 \sqrt{x}}$. Applying the chain rule gives
$$\frac{d}{dx} \ln(\sqrt{x}) = \frac{1}{\sqrt{x}}\frac{1}{2 \sqrt{x}} = \frac{1}{2 \sqrt{x}^2}$$
and since $x > 0$, this simplifies to $\frac{1}{2x}$.
