Question

If \[y = \tan^{- 1} \left\{ \frac{\log_e \left( e/ x^2 \right)}{\log_e \left( e x^2 \right)} \right\} + \tan^{- 1} \left( \frac{3 + 2 \log_e x}{1 - 6 \log_e x} \right)\], then \[\frac{d^2 y}{d x^2} =\]

 

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Answer 1

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\[y = \tan^{- 1} \left\{ \frac{\log_e \left( e/ x^2 \right)}{\log_e \left( e x^2 \right)} \right\} + \tan^{- 1} \left( \frac{3 + 2 \log_e x}{1 - 6 \log_e x} \right)\]

\[ \Rightarrow y = \tan^{- 1} \left\{ \frac{1 - 2 \log_e x}{1 + 2 \log_e x} \right\} + \tan^{- 1} \left( \frac{3 + 2 \log_e x}{1 - 6 \log_e x} \right)\]

\[ \Rightarrow y = \tan^{- 1} \left( \frac{\frac{1 - 2 \log_e x}{1 + 2 \log_e x} + \frac{3 + 2 \log_e x}{1 - 6 \log_e x}}{1 - \left( \frac{1 - 2 \log_e x}{1 + 2 \log_e x} \right)\left( \frac{3 + 2 \log_e x}{1 - 6 \log_e x} \right)} \right)\]

\[ \Rightarrow y = \tan^{- 1} \left\{ \frac{\left( 1 - 2 \log_e x \right)\left( 1 - 6 \log_e x \right) + \left( 3 + 2 \log_e x \right)\left( 1 + 2 \log_e x \right)}{\left( 1 + 2 \log_e x \right)\left( 1 - 6 \log_e x \right) - \left( 1 - 2 \log_e x \right)\left( 3 + 2 \log_e x \right)} \right\}\]

\[ \Rightarrow y = \tan^{- 1} \left\{ \frac{1 - 8 \log_e x + 12 \left( \log_e x \right)^2 + 3 + 8 \log_e x + 4 \left( \log_e x \right)^2}{1 - 4 \log_e x - 12 \left( \log_e x \right)^2 - 3 + 4 \log_e x + 4 \left( \log_e x \right)^2} \right\}\]

\[ \Rightarrow y = \tan^{- 1} \left\{ \frac{1 - 8 \log_e x + 12 \left( \log_e x \right)^2 + 3 + 8 \log_e x + 4 \left( \log_e x \right)^2}{1 - 4 \log_e x - 12 \left( \log_e x \right)^2 - 3 + 4 \log_e x + 4 \left( \log_e x \right)^2} \right\}\]

\[ \Rightarrow y = \tan^{- 1} \left\{ \frac{4 + 16 \left( \log_e x \right)^2}{- 2 - 8 \left( \log_e x \right)^2} \right\}\]

\[ \Rightarrow y = \tan^{- 1} \left[ \frac{4\left\{ 1 + 4 \left( \log_e x \right)^2 \right\}}{- 2\left\{ 1 + 4 \left( \log_e x \right)^2 \right\}} \right]\]

\[ \Rightarrow y = \tan^{- 1} \left[ - 2 \right]\]

\[ \Rightarrow \frac{dy}{dx} = 0\]

\[ \Rightarrow \frac{d^2 y}{d x^2} = 0\]