Limits, Continuity and Differentiability
80271
If \(y=\tan ^{-1}\left[\frac{\log (e x)}{\log \left(\frac{e}{x}\right)}\right]\), then \(\frac{d y}{d x}=\)
1 \(\frac{-x}{1+(\log x)^{2}}\)
2 \(\frac{x}{1+(\log x)^{2}}\)
3 \(\frac{1}{x\left[1+(\log x)^{2}\right]}\)
4 \(\frac{-1}{x\left[1+(\log x)^{2}\right]}\)
Explanation:
(C) : Given,
We know that,
\(y=\tan ^{-1}\left[\frac{\log (e x)}{\log \left(\frac{e}{x}\right)}\right]\)
\(\log (a b)=\log a+\log b\)
\(\log \left(\frac{a}{b}\right)=\log a-\log b\)
Then,
\(y=\tan ^{-1}\left[\frac{\log e+\log x}{\log e-\log x}\right]\)
\(y=\tan ^{-1}\left[\frac{1+\log x}{1-\log x}\right]\)
Let, \(\quad \log x=\tan \theta\)
\(\theta=\tan ^{-1}(\log x)\)
Then, \(y=\tan ^{-1}\left[\frac{1+\tan \theta}{1-\tan \theta}\right]\)
\(y=\tan ^{-1}\left[\frac{\tan \frac{\pi}{4}+\tan \theta}{1-\tan \frac{\pi}{4} \cdot \tan \theta}\right] \Rightarrow y=\tan ^{-1}\left[\tan \left(\frac{\pi}{4}+\theta\right)\right]\)
\(\mathrm{y}=\frac{\pi}{4}+\theta\)
Putting the value of \(\theta\) in above equation-
\(y=\frac{\pi}{4}+\tan ^{-1}(\log x)\)
\(\therefore \quad \frac{\mathrm{dy}}{\mathrm{dx}}=0+\frac{1}{1+(\log \mathrm{x})^{2}} \frac{\mathrm{d}}{\mathrm{dx}}(\log \mathrm{x})\)
\(\frac{d y}{d x}=\frac{1}{x\left[1+(\log x)^{2}\right]}\)