Given $y=f(x)$, where $f(x) $ is a positive function, we can write $\ln y = \ln f(x) $. Now let's say that $f$ takes zero values at certain points in an interval. At these points, the natural logarithm of the function is not defined. Take the example of $\sin(x) +1$ in $[\pi, 2\pi]$. It takes zero value at $3\pi /2 $. At this point, the tangent is horizontal, we see. We, however, cannot determine the slope of this tangent by doing logarithmic differentiation because the derivative at this point is indeterminate.
I have come across a problem that asks me to use logarithmic differentiation to evaluate the derivative of $\sqrt{\frac{t}{t+1}}$. Here, $t=0$ is in the domain of $y$ and not in the domain of $\ln y$. How is this logarithmic differentiable?
Another problem asks me to evaluate the derivative of $\tan(x) \sqrt{2x +1}$ using logarithmic differentiation. The domain consists of all $x \geq - 1/2$, and $x\neq (2n+1)\pi /2$, where $n$ is an integer. For several $x$ in the domain, the function takes negative values at several points. How can this be logarithmic differentiable?
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$\begingroup$Normally you don't need logarithmic differentiation unless you get an expression of the form $\{f(x) \}^{g(x)} $ where both $f, g$ are non-constant. Since $g$ is expected to be differentiable, it is also continuous and further being non-constant it does take irrational values and then the expression $f^{g} $ is defined only when $f$ is positive and the whole expression is positive.
Any other use of logarithmic differentiation (like for complicated products and quotients or radical expression) can be proved equivalent to the usual rules of differentiation (product/quotient/chain rule). So one should understand that in such cases the use of logarithmic differentiation is more of technical device to simplify calculation and the same result can be achieved with other methods also.
It is better to illustrate my point with an actual example. Consider the usual product rule $$(uv) '=uv' +u'v$$ and dividing this with $uv$ we get $$\frac{(uv) '} {uv} =\frac{u'} {u} +\frac{v'} {v} $$ or $$(\log uv)' =(\log u) '+(\log v)' $$ which is expected because $\log uv=\log u +\log v$. Here you can see that the logarithm does not give us anything new except for just a simple mechanism to remember product rule. Same is the case with quotient rule or chain rule used for radicals. And it does not matter whether the argument of log is positive or not.
$\endgroup$ 15 $\begingroup$for your first example: $$\ln(y)=\frac{1}{2}\ln\left(\frac{t}{t+1}\right)=\frac{1}{2}\left(\ln(t)-\ln(t+1)\right)$$ then $$\frac{y'}{y}=\left(\frac{1}{t}-\frac{1}{t+1}\right)$$ for the function $$y=\sqrt{\frac{t}{t+1}}$$ and $$t\geq 0$$ is the first derivative $$y'=\frac{1}{2}\left(\sqrt{\frac{t}{t+1}}\right)^{-1/2}\cdot \frac{1}{(t+1)^2}$$ here must also be $$t>0$$
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