Question
If $$y = {\log ^n}x,$$ where $${\log ^n}$$ means $$\log \,\log \,\log .....$$ (repeated $$n$$ time), then $$x\,\log \,x\,\log \,x\,{\log ^2}x\,{\log ^3}x.....{\log ^{n - 1}}x\,{\log ^n}x\frac{{dy}}{{dx}}$$ is equal to :
A.
$$\log \,x$$
B.
$${\log ^n}x$$
C.
$$\frac{1}{{\log \,x}}$$
D.
$$1$$
Answer :
$${\log ^n}x$$
Solution :
$$\because \,y = {\log ^n}x$$
On differentiating w.r.t. $$x,$$ we get
$$\eqalign{
& x\,\log \,x\,{\log ^2}x\,{\log ^3}x.....{\log ^{n - 1}}x\,{\log ^n}x\frac{{dy}}{{dx}} \cr
& = \frac{{x\,\log \,x\,{{\log }^2}x\,{{\log }^3}x.....{{\log }^{n - 1}}x\,{{\log }^n}x.1}}{{x\,\log \,x\,{{\log }^2}x\,{{\log }^3}x.....{{\log }^{n - 1}}x}} \cr
& = {\log ^n}x \cr} $$