If f(x) = tan^{-1}left{ frac{log(e/x^2)}{log( | vedclass

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(B) We have $y = 	an^{-1}\left( \frac{\log e - \log x^2}{\log e + \log x^2} ight) + 	an^{-1}\left( \frac{3 + 2\log x}{1 - 6\log x} ight)$.Using

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(B) We have $y = 	an^{-1}\left( \frac{\log e - \log x^2}{\log e + \log x^2} ight) + 	an^{-1}\left( \frac{3 + 2\log x}{1 - 6\log x} ight)$.Using the properties of logarithms,$\log e = 1$ and $\log x^2 = 2\log x$,so:$y = 	an^{-1}\left( \frac{1 - 2\log x}{1 + 2\log x} ight) + 	an^{-1}\left( \frac{3 + 2\log x}{1 - 6\log x} ight)$.Applying the formula $	an^{-1} A - 	an^{-1} B = 	an^{-1}\left( \frac{A-B}{1+AB} ight)$:$y = (	an^{-1} 1 - 	an^{-1}(2\log x)) + (	an^{-1} 3 + 	an^{-1}(2\log x))$.Simplifying the expression:$y = 	an^{-1} 1 + 	an^{-1} 3$.Since $y$ is a constant,its derivative with respect to $x$ is zero:$\frac{dy}{dx} = 0$.Consequently,for any $n \ge 1$,the $n$-th order derivative is:$\frac{d^n y}{dx^n} = 0$.

If $f(x) = \tan^{-1}\left\{ \frac{\log(e/x^2)}{\log(ex^2)} \right\} + \tan^{-1}\left( \frac{3 + 2\log x}{1 - 6\log x} \right)$,then $\frac{d^n y}{dx^n}$ is $(n \ge 1)$.

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