If ` y= sin^(-1)"" (2x )/(1+x^2) + sec^(-1)"" (1+x^2)/(1-x^2) `, then ` (dy)/(dx)=`

Options

A. `(4)/( 1+x^2)`
B. `(2)/( 1+x^2)`
C. ` (3)/( 1+x^2)`
D. None

Answer

To find the derivative \( \frac{dy}{dx} \) of the function

\[ y = \sin^{-1}\left(\frac{2x}{1+x^2}\right) + \sec^{-1}\left(\frac{1+x^2}{1-x^2}\right), \]

we will follow these steps:

### Step 1: Simplify the function using trigonometric identities

We can use the substitution \( x = \tan(\theta) \). Thus, we have:

\[ \sin^{-1}\left(\frac{2\tan(\theta)}{1+\tan^2(\theta)}\right) = \sin^{-1}(\sin(2\theta)) = 2\theta, \]

because \( \frac{2\tan(\theta)}{1+\tan^2(\theta)} \) is the double angle formula for sine.

### Step 2: Simplify the second term

For the second term, we have:

\[ \sec^{-1}\left(\frac{1+\tan^2(\theta)}{1-\tan^2(\theta)}\right) = \sec^{-1}(\sec(2\theta)) = 2\theta, \]

since \( \frac{1+\tan^2(\theta)}{1-\tan^2(\theta)} \) is the double angle formula for secant.

### Step 3: Combine the results

Now, we can combine the results from the two terms:

\[ y = 2\theta + 2\theta = 4\theta. \]

Since \( \theta = \tan^{-1}(x) \), we can rewrite \( y \) as:

\[ y = 4\tan^{-1}(x). \]

### Step 4: Differentiate \( y \)

Now we differentiate \( y \) with respect to \( x \):

\[ \frac{dy}{dx} = 4 \cdot \frac{d}{dx}(\tan^{-1}(x)). \]

Using the derivative of \( \tan^{-1}(x) \), which is \( \frac{1}{1+x^2} \):

\[ \frac{dy}{dx} = 4 \cdot \frac{1}{1+x^2}. \]

### Final Result

Thus, the final result is:

\[ \frac{dy}{dx} = \frac{4}{1+x^2}. \]

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