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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?

R004
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2 Answers2

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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.

Paramanand Singh
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  • Don't you think that it's better to write the anti derivative of $\frac{u'}{u}$ as $\ln \left | u \right |$? If $u$ is a negative function, the last $\ln u$ you have put forward will not make sense, right? – R004 Jul 24 '17 at 14:00
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    @R004 : Perhaps you did not get my point. The above argument shows that logarithmic differentiation can be used as a technique to help remember the product/quotient/chain rule by just keeping track of the properties of logarithm function. Actual answer for the derivative of $uv$ is still $u'v+uv'$ and hence we don't need to actually worry about signs. – Paramanand Singh Jul 24 '17 at 16:28
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    @R004 : for a beginner the product rule for derivatives may seem bit unnatural (the natural expectation is $(uv) '=u' v'$ similar to $(u+v) '=u' +v'$). But once you know properties of logarithm function, you can see that nothing could be more natural or expected than the product rule for derivatives. Similar remarks apply to quotient rule and chain rule ($(u^{n}) '=nu^{n-1}u'$). – Paramanand Singh Jul 24 '17 at 16:33
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    @R004 : You must understand that the above argument cannot be a proof for product rule precisely because it applies only when the functions are taking positive values whereas the product rule holds without this constraint also. The argument only tries to make the form of product / quotient rule less mysterious and perhaps easy to remember (in the sense that you take logs do the derivatives and simplify without any need to remember cryptic product and quotient rule). You just need to remember derivative of log and the common properties of logarithm. – Paramanand Singh Jul 24 '17 at 16:38
  • Can I interpret this in the following way? We have, $y=f(x)$. Taking logarithm on both sides we get, $\ln y= \ ln f(x) = \int_{1}^{f(x)} \frac{1}{t} dt$. Taking the derivative on both sides would be applying Liebnitz's rule, a rule that only requires $f(x)$ to be differentiable at all x( and $1$ is of course differentiable ) and $\frac{1}{t}$ to be defined at all $f(x)$( and it is of course defined at $1$ ). This rule does not talk about the necessity of a function to be positive or negative. – R004 Jul 25 '17 at 01:14
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    @R004 : the integral in your comment does not make sense if $f(x) \leq 0$. Try whatever way you think of the limitation of positive values remains with the logarithm function . The logarithm function can be used to figure out the exact form of product and quotient rule and nothing more than that. If you stick to single variable calculus there is no rule for differentiation of $u^{v} $ and this case necessitates the use of logarithm function. This is also reasonable to expect because one of the popular definitions of symbol $u^{v} $ is $\exp(v\log u) $. – Paramanand Singh Jul 25 '17 at 02:23
  • I get this. This helps. – R004 Jul 25 '17 at 03:45
  • I couldn't help but think more about this. Let $y=f(x)$. Then, $\ln \left | y \right | = \ln \left | f(x) \right |$, where $y>0$. On differentiating bother sides we get, $\frac{1}{\left | y \right |} \frac{\left | y \right |}{y} y' = \frac{1}{\left | f(x) \right |} \frac{\left | f(x) \right |}{f(x)} f'$. This gives, $y'=f'$ for any positive or negative function. – R004 Jul 27 '17 at 05:16
  • @R004 : problem with taking mods is that $|f|$ may be differentiable even when $f$ is not. And this does not deal with points where $f$ vanishes. – Paramanand Singh Jul 27 '17 at 05:26
  • Sorry. $\left | y \right | > 0$ as per the definition of natural logarithm. And yes, it does not tell us anything when $y=0$. – R004 Jul 27 '17 at 05:33
  • Any example in which the absolute function is differentiable when the function itself is not? – R004 Jul 27 '17 at 05:37
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    @R004 : take $f(x) =1$ if $x$ is rational and $f(x) =-1$ when $x$ is irrational. Then $f$ is discontinuous everywhere and $|f|=1$ everywhere. – Paramanand Singh Jul 27 '17 at 07:13
  • I must add this. As per the definition of $\ln$, $\left | y \right | > 0$. This implies that $y\neq0$. The explanation holds. Although $y$ may take zero values, the natural logarithm restricts – R004 Jul 27 '17 at 07:17
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    @R004 : Perhaps you did not get it. Taking absolute value is an irreversible step. There is no way to get $f$ from $|f|$ and using results about $|f|$ to deduce anything about $f$ is logically impossible. – Paramanand Singh Jul 27 '17 at 07:21
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    The function you have given is infinitely discontinuous. It is not differentiable. This, I think, restricts the use of logarithmic differentiation to differentiable functions. – R004 Jul 27 '17 at 07:21
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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 $$t>0$$

Dr. Sonnhard Graubner
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    Let $g = \ln y$. We see that $g$ is undefined at $t=0$. If a function is undefined at a point, how can it be differentiable at that point? The math does bring me to a point where I can agree with the result. But getting started with a function taking a zero value is not comforting. – R004 Jul 24 '17 at 10:21