If f(x) = (log _{cot x}tan x)(log _{tan x}cot | vedclass

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(B) Given $f(x) = (\log _{\cot x}	an x)(\log _{	an x}\cot x)^{-1}$.We know that $\log _{a}b = \frac{1}{\log _{b}a}$.Let $u = \log _{\cot x}	an x$.

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(B) Given $f(x) = (\log _{\cot x}	an x)(\log _{	an x}\cot x)^{-1}$.We know that $\log _{a}b = \frac{1}{\log _{b}a}$.Let $u = \log _{\cot x}	an x$. Then $\log _{	an x}\cot x = \frac{1}{u}$.Substituting this into the expression for $f(x)$:$f(x) = u \cdot (u^{-1})^{-1} = u \cdot u = u^2$.Since $	an x = (\cot x)^{-1}$,we have $u = \log _{\cot x}(\cot x)^{-1} = -1 \cdot \log _{\cot x}(\cot x) = -1$.Thus,$f(x) = (-1)^2 = 1$.Since $f(x) = 1$ is a constant function,its derivative $f'(x) = 0$ for all $x$ in the domain.Therefore,$f'(2) = 0$.

If $f(x) = (\log _{\cot x}\tan x)(\log _{\tan x}\cot x)^{-1}$,then $f'(2) = $

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