Limits, Continuity and Differentiability
80316
If \(y=\tan ^{-1}\left(\frac{4 x}{1+5 x^{2}}\right)+\cot ^{-1}\left(\frac{3-2 x}{2+3 x}\right)\), then \(\frac{d y}{d x}\)
is
1 \(\frac{-5}{1+25 \mathrm{x}^{2}}\)
2 \(\frac{5}{1+25 \mathrm{x}^{2}}\)
3 \(\frac{5}{1-25 \mathrm{x}^{2}}\)
4 0
Explanation:
(B) : Given,
\(y=\tan ^{-1}\left(\frac{4 x}{1+5 x^{2}}\right)+\cot ^{-1}\left(\frac{3-2 x}{2+3 x}\right)\)
\(y=\tan ^{-1}\left(\frac{5 x-x}{1+5 x^{2}}\right)+\cot ^{-1}\left(\frac{\frac{3}{2}-x}{1+\frac{3}{2} x}\right)\)
We know that,
\(\tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}\)
\(\cot ^{-1} x=\frac{\pi}{2}-\tan ^{-1} x\)
Then,
\(y=\tan ^{-1}\left(\frac{5 x-x}{1+5 x^{2}}\right)+\frac{\pi}{2}-\tan ^{-1}\left(\frac{\frac{3}{2}-x}{1+\frac{3}{2} x}\right)\)
Again we know that,
Then,
\(\tan ^{-1} a-\tan ^{-1} b=\tan ^{-1}\left(\frac{a-b}{1+a b}\right)\)
\(y=\left(\tan ^{-1} 5 x-\tan ^{-1} x\right)+\frac{\pi}{2}-\left(\tan ^{-1} \frac{3}{2}-\tan ^{-1} x\right)\)
\(y=\tan ^{-1} 5 x-\tan ^{-1} \frac{3}{2}+\tan ^{-1} x-\tan ^{-1} x+\frac{\pi}{2}\)
\(y=\tan ^{-1} 5 x-\tan ^{-1} \frac{3}{2}+\frac{\pi}{2}\)
\(\therefore \quad \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{5}{1+(5 \mathrm{x})^{2}}-0+0 \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{5}{1+25 \mathrm{x}^{2}}\)