If `u=sin^(-1)((2x)/(1+x^(2)))andv=tan^(-1)((2x)/(1-x^(2)))`, then `(du)/(dv)` is

Options

A. `(1)/(2)`
B. 2x
C. `(1-x^(2))/(1+x^(2))`
D. 1

Answer

To find \(\frac{du}{dv}\) given \(u = \sin^{-1}\left(\frac{2x}{1+x^2}\right)\) and \(v = \tan^{-1}\left(\frac{2x}{1-x^2}\right)\), we can follow these steps:

### Step 1: Express \(u\) and \(v\) in terms of \(\theta\)

Let \(x = \tan(\theta)\). Then we can rewrite \(u\) and \(v\) using trigonometric identities.

- For \(u\):
\[ u = \sin^{-1}\left(\frac{2\tan(\theta)}{1+\tan^2(\theta)}\right) \]
Using the identity \(\frac{2\tan(\theta)}{1+\tan^2(\theta)} = \sin(2\theta)\), we have:
\[ u = \sin^{-1}(\sin(2\theta)) = 2\theta \]

- For \(v\):
\[ v = \tan^{-1}\left(\frac{2\tan(\theta)}{1-\tan^2(\theta)}\right) \]
Using the identity \(\frac{2\tan(\theta)}{1-\tan^2(\theta)} = \tan(2\theta)\), we have:
\[ v = \tan^{-1}(\tan(2\theta)) = 2\theta \]

### Step 2: Differentiate \(u\) and \(v\)

Now, we differentiate \(u\) and \(v\) with respect to \(\theta\):

- The derivative of \(u\) with respect to \(\theta\):
\[ \frac{du}{d\theta} = 2 \]

- The derivative of \(v\) with respect to \(\theta\):
\[ \frac{dv}{d\theta} = 2 \]

### Step 3: Find \(\frac{du}{dv}\)

Now we can find \(\frac{du}{dv}\):
\[ \frac{du}{dv} = \frac{du/d\theta}{dv/d\theta} = \frac{2}{2} = 1 \]

### Final Result

Thus, the value of \(\frac{du}{dv}\) is:
\[ \frac{du}{dv} = 1 \]