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If \[y = {\log ^n}x\] where \[{\log ^n}\] means \[\log .\log .\log \] repeated \[n\] times, then find the value of \[x.\log x.{\log ^2}x.{\log ^3}x - - - {\log ^{n - 1}}x.\dfrac{{dy}}{{dx}} = \]

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Last updated date: 23rd Jul 2024
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Answer
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Hint: Here we are given to find the value of differentiation of log function and then find the product of various log functions and their differentiation. To do this we first find the differentiation of log for powers two and three. Then we use patterns formed in those values to get the differentiation of log for power \[n\] . Then we find the value of the given expression by putting the value of differentiation of log for power \[n\] in it.
Formula used: We do the differential of \[\dfrac{{da(b(x))}}{{dx}}\] as
\[\dfrac{{da(b(x))}}{{dx}} = \dfrac{{da(b(x))}}{{db(x)}} \cdot \dfrac{{db(x)}}{{dx}}\]
And
\[\dfrac{{d\log x}}{{dx}} = \dfrac{1}{x}\]

Complete step-by-step solution:
We are given \[y = {\log ^n}x\]. We have to find the value of \[\dfrac{{dy}}{{dx}}\]. Since it is lengthy to find the differentiation directly for power \[n\], we first find it for,
\[n = 2\],
\[y = {\log ^2}x\]
\[
   \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{d{{\log }^2}x}}{{dx}} \\
   \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{d\log (\log x)}}{{dx}} \\
 \]
We know that, \[\dfrac{{d\log x}}{{dx}} = \dfrac{1}{x}\]and \[\dfrac{{da(b(x))}}{{dx}} = \dfrac{{da(b(x))}}{{db(x)}} \cdot \dfrac{{db(x)}}{{dx}}\], using this we get,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{\log x}} \cdot \dfrac{1}{x}\]
Same for \[n = 3\], we get
\[
  y = {\log ^3}x \\
   \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{d{{\log }^3}x}}{{dx}} \\
   \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{d\log (\log (\log x))}}{{dx}} \\
 \]
Using \[\dfrac{{d\log x}}{{dx}} = \dfrac{1}{x}\] and \[\dfrac{{da(b(x))}}{{dx}} = \dfrac{{da(b(x))}}{{db(x)}} \cdot \dfrac{{db(x)}}{{dx}}\] ,we get
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{{{\log }^2}x}}\dfrac{1}{{\log x}} \cdot \dfrac{1}{x}\]
Same way we can say for \[n\] as,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{{{\log }^n}x}} \cdot \dfrac{1}{{{{\log }^{n - 1}}x}}.......\dfrac{1}{{\log x}} \cdot \dfrac{1}{x}\]
Now using this we get the value of \[x.\log x.{\log ^2}x.{\log ^3}x - - - {\log ^{n - 1}}x.\dfrac{{dy}}{{dx}}\] as,
\[
   \Rightarrow x.\log x.{\log ^2}x.{\log ^3}x - - - {\log ^{n - 1}}x.\dfrac{{dy}}{{dx}} = \left( {x.\log x.{{\log }^2}x.{{\log }^3}x - - - {{\log }^{n - 1}}x} \right)\left( {\dfrac{1}{{{{\log }^n}x}} \cdot \dfrac{1}{{{{\log }^{n - 1}}x}}.......\dfrac{1}{{\log x}} \cdot \dfrac{1}{x}} \right) \\
   \Rightarrow x.\log x.{\log ^2}x.{\log ^3}x - - - {\log ^{n - 1}}x.\dfrac{{dy}}{{dx}} =\dfrac{1}{{\log ^n}x} \]
Hence we get the answer as \[\dfrac{1}{{\log ^n}x}\].

Note: This is to note that here we have done the differentiation of a function having any general power using the method of trial. We have found a pattern in differentiation by putting some smaller powers. Had a pattern been not found, we would have used any other method here. We could have also found the differentiation directly, but it required bigger calculations. Hence we have saved some time here. We should always look for such alternate methods as they make calculations easier.