Question

If \[y = \log{^n}x\], where \[\log{ ^n}\] means \[\log \log \log....\](repeated \[n\] times). Then what is the value of \[xlogx\log{^2}x\log{^3}x...\log{^{n - 1}}x\log{^n}x\dfrac{{dy}}{{dx}}\]?
A. \[\log x\]
B. \[x\]
C. \[\dfrac{1}{{\log x}}\]
D. \[\log{^n}x\]

Answer 1

Hint: In the given question, one logarithmic equation is given. Substitute various values of \[n\] in the given equation and simplify it by using the given logarithmic condition. Then differentiate the equation with respect to \[x\]. By following this step up to \[n = n\], we will find the value of \[x\log x\log{ ^2}x\log{ ^3}x...\log{ ^{n - 1}}x\log{^n}x\dfrac{{dy}}{{dx}}\].
Formula Used:
\[\dfrac{d}{{dx}}\left( {\log x} \right) = \dfrac{1}{x}\]
\[\dfrac{d}{{dx}}\left( {\log \log x} \right) = \dfrac{1}{{\log x}}\dfrac{d}{{dx}}\left( {\log x} \right)\]
Complete step by step solution:
The given \logarithmic equation is \[y = \log^{n}x\], where \[\log^{n} = \log \log \log....\](repeated \[n\] times)
Let’s simplify the given equation.
Substitute \[n = 2\] in the above equation.
\[y = \log^{2}x\]
Apply the given condition \[\log^{n} = \log \log \log....\](repeated \[n\] times).
Then,
\[y = \log \log x\]
Differentiate the above equation with respect to \[x\].
\[\dfrac{{dy}}{{dx}} = \dfrac{1}{{\log x}}\dfrac{d}{{dx}}\left( {\log x} \right)\]
\[ \Rightarrow \]\[\dfrac{{dy}}{{dx}} = \dfrac{1}{{\log x}}\left( {\dfrac{1}{x}} \right)\]
\[ \Rightarrow \]\[\dfrac{{dy}}{{dx}} = \dfrac{1}{{x\log x}}\]

Now substitute \[n = 3\] in the above equation.
\[y = \log^{3}x\]
Apply the given condition \[\log^{n} = \log \log \log....\](repeated \[n\] times).
Then,
\[y = \log \log \log x\]
Differentiate the above equation with respect to \[x\].
\[\dfrac{{dy}}{{dx}} = \dfrac{1}{{\log \log x}}\dfrac{d}{{dx}}\left( {\log \log x} \right)\]
\[ \Rightarrow \]\[\dfrac{{dy}}{{dx}} = \dfrac{1}{{\log \log x}}\left[ {\dfrac{1}{{\log x}}\dfrac{d}{{dx}}\left( {\log x} \right)} \right]\]
\[ \Rightarrow \]\[\dfrac{{dy}}{{dx}} = \dfrac{1}{{\log \log x}}\left[ {\dfrac{1}{{\log x}}\left( {\dfrac{1}{x}} \right)} \right]\]
\[ \Rightarrow \]\[\dfrac{{dy}}{{dx}} = \dfrac{1}{{x\log x\log^{2}x}}\]

Similarly substitute \[n = n\] in the above equation.
\[y = \log^{n}x\]
Apply the given condition \[\log^{n} = \log \log \log....\](repeated \[n\] times).
Then,
\[y = \log \log \log....(repeated\,\, n - 1\,\, times) \log x\]
Differentiate the above equation with respect to \[x\] and use the pattern followed by the derivative of above equations.
\[\dfrac{{dy}}{{dx}} = \dfrac{1}{{\log x \log^{2}x\log^{3}x.....\log^{n - 1}x}}\left( {\dfrac{1}{x}} \right)\]
\[ \Rightarrow \]\[\dfrac{{dy}}{{dx}} = \dfrac{1}{{x\log x \log^{2}x\log^{3}x.....\log^{n - 1}}x}\]
Simplify the above differential equation.
\[x\log x \log^{2}x\log^{3}x.....\log^{n - 1}x\dfrac{{dy}}{{dx}} = 1\]
Now multiply both sides of above equation by \[\log^{n}x\].
Then,
\[x\log x \log^{2}x\log^{3}x.....\log{n - 1}x\log^{n}x\dfrac{{dy}}{{dx}} = \log^{n}x\]

Hence the correct option is option D.
Note: Students are often get confused with the differentiation of multiple logarithmic or iterative logarithmic functions. In this problem, to find the derivative of iterative logarithmic function, we have to use the product rule of derivative \[\dfrac{d}{{dx}}\left( {uv} \right) = u\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}}\].