KCET · Maths · Differentiation
If \(u=\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)\) and \(v=\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)\) then \(\frac{d u}{d v}\) is
- A \(2\)
- B \(\frac{1-x^2}{1+x^2}\)
- C \(1\)
- D \(\frac{1}{2}\)
Answer & Solution
Correct Answer
(C) \(1\)
Step-by-step Solution
Detailed explanation
Here, \(u=\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)\)
\(\left.u=2 \tan ^{-1} x \quad \because \because 2 \tan ^{-1} x=\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)\right]\)
\(\begin{aligned} & \Rightarrow \frac{d u}{d x}=2 \times \frac{1}{1+x^2} \\ & \text { and } v=\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)\end{aligned}\)
\(\begin{aligned} y & =2 \tan ^{-1} x \quad\left[\because 2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)\right] \\ \Rightarrow \quad & \quad \frac{d v}{d x}=\frac{2}{1+x^2}\end{aligned}\)
From Eqs. (i) and (ii), we get
\(\frac{d v}{d x}=1\)
\(\left.u=2 \tan ^{-1} x \quad \because \because 2 \tan ^{-1} x=\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)\right]\)
\(\begin{aligned} & \Rightarrow \frac{d u}{d x}=2 \times \frac{1}{1+x^2} \\ & \text { and } v=\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)\end{aligned}\)
\(\begin{aligned} y & =2 \tan ^{-1} x \quad\left[\because 2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)\right] \\ \Rightarrow \quad & \quad \frac{d v}{d x}=\frac{2}{1+x^2}\end{aligned}\)
From Eqs. (i) and (ii), we get
\(\frac{d v}{d x}=1\)
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