If y=tan−1⁡log⁡ex2log⁡ex2+tan−1⁡3+2log⁡x1−6log⁡x then dydxx−2+dydxx−3=

y=tan−1⁡1−2log⁡x1+2log⁡x+tan−1⁡3+2log⁡x1−6log⁡x=tan−1⁡1−2log⁡x1+2log⁡x+3+2log⁡x1−6log⁡x1−1−2log⁡x1+2log⁡x3+2log⁡x1−6log⁡x=tan−1⁡(−2)⇒dydx=0⇒dydxx=2+dydxx=3=0

If y=tan−1⁡log⁡ex2log⁡ex2+tan−1⁡3+2log⁡x1−6log⁡x then dydxx−2+dydxx−3=

y=tan−1⁡1−2log⁡x1+2log⁡x+tan−1⁡3+2log⁡x1−6log⁡x=tan−1⁡1−2log⁡x1+2log⁡x+3+2log⁡x1−6log⁡x1−1−2log⁡x1+2log⁡x3+2log⁡x1−6log⁡x=tan−1⁡(−2)⇒dydx=0⇒dydxx=2+dydxx=3=0

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$If y = \tan^{- 1} (\frac{\log (\frac{e}{x^{2}})}{\log e x^{2}}) + \tan^{- 1} (\frac{3 + 2 \log x}{1 - 6 \log x}) then {(\frac{d y}{d x})}_{x - 2} + {(\frac{d y}{d x})}_{x - 3} =$

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Detailed Solution

$\begin{array}{l} y = \tan^{- 1} (\frac{1 - 2 \log x}{1 + 2 \log x}) + \tan^{- 1} (\frac{3 + 2 \log x}{1 - 6 \log x}) = \tan^{- 1} [\frac{\frac{1 - 2 \log x}{1 + 2 \log x} + \frac{3 + 2 \log x}{1 - 6 \log x}}{1 - (\frac{1 - 2 \log x}{1 + 2 \log x}) (\frac{3 + 2 \log x}{1 - 6 \log x})}] = \tan^{- 1} (- 2) \\ \Rightarrow \frac{d y}{d x} = 0 \Rightarrow {(\frac{d y}{d x})}_{x = 2} + {(\frac{d y}{d x})}_{x = 3} = 0 \end{array}$