If y = tan^{-1} left( frac{log(e/x^3)}{log(ex | vedclass

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(B) Let $u = \log x$. Then $\log(e/x^3) = 1 - 3u$ and $\log(ex^3) = 1 + 3u$.Also, $\log(e^4x^3) = 4 + 3u$ and $\log(e/x^{12}) = 1 - 12u$.Substituting

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(B) Let $u = \log x$. Then $\log(e/x^3) = 1 - 3u$ and $\log(ex^3) = 1 + 3u$.Also, $\log(e^4x^3) = 4 + 3u$ and $\log(e/x^{12}) = 1 - 12u$.Substituting these into the expression for $y$:$y = 	an^{-1} \left( \frac{1-3u}{1+3u} ight) + 	an^{-1} \left( \frac{4+3u}{1-12u} ight)$.Using the identity $	an^{-1} A + 	an^{-1} B = 	an^{-1} \left( \frac{A+B}{1-AB} ight)$:$y = 	an^{-1} \left( \frac{\frac{1-3u}{1+3u} + \frac{4+3u}{1-12u}}{1 - \left( \frac{1-3u}{1+3u} ight) \left( \frac{4+3u}{1-12u} ight)} ight)$.Simplifying the numerator: $(1-3u)(1-12u) + (4+3u)(1+3u) = (1 - 15u + 36u^2) + (4 + 15u + 9u^2) = 5 + 45u^2$.Simplifying the denominator: $(1+3u)(1-12u) - (1-3u)(4+3u) = (1 - 9u - 36u^2) - (4 - 9u - 9u^2) = -3 - 27u^2 = -3(1 + 9u^2)$.This approach suggests $y$ is a constant or simplifies to a specific value. Re-evaluating: $	an^{-1} \left( \frac{1-3u}{1+3u} ight) = 	an^{-1}(1) - 	an^{-1}(3u) = \frac{\pi}{4} - 	an^{-1}(3u)$.And $	an^{-1} \left( \frac{4+3u}{1-12u} ight) = 	an^{-1}(4) + 	an^{-1}(3u)$.Thus, $y = \frac{\pi}{4} + 	an^{-1}(4)$, which is a constant.Therefore, $\frac{dy}{dx} = 0$.

If $y = \tan^{-1} \left( \frac{\log(e/x^3)}{\log(ex^3)} \right) + \tan^{-1} \left( \frac{\log(e^4x^3)}{\log(e/x^{12})} \right)$, for $x \in (e^{-1/3}, e^{1/12})$, then $\frac{dy}{dx}$ is equal to...

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