Derivative of ` tan ^(-1) ((sqrt( 1+x^(2))-1)/( x)) w.r.t. tan ^(-1) ((2x sqrt(1-x^(2)))/( 1-2x ^(2)))` is

Options

A. ` (-sqrt( 1-x^(2)))/( 4( 1+x^(2)))`
B. ` (sqrt( 1-x^(2)))/( 4( 1+x^(2)))`
C. ` (-sqrt( 1-x^(2)))/(1+x^(2))`
D. ` ( sqrt( 1-x^(2)))/( 1+x^(2))`

Answer

To find the derivative of \( f(x) = \tan^{-1} \left( \frac{\sqrt{1+x^2} - 1}{x} \right) \) with respect to \( g(x) = \tan^{-1} \left( \frac{2x \sqrt{1-x^2}}{1-2x^2} \right) \), we will follow these steps:

### Step 1: Simplify \( f(x) \)

We start with:
\[ f(x) = \tan^{-1} \left( \frac{\sqrt{1+x^2} - 1}{x} \right) \]

Let \( x = \tan(\theta) \). Then, we have:
\[ \sqrt{1+x^2} = \sqrt{1+\tan^2(\theta)} = \sec(\theta) \]
Thus,
\[ f(x) = \tan^{-1} \left( \frac{\sec(\theta) - 1}{\tan(\theta)} \right) \]

Now, we can rewrite the expression:
\[ \frac{\sec(\theta) - 1}{\tan(\theta)} = \frac{\sec(\theta) - 1}{\frac{\sin(\theta)}{\cos(\theta)}} = \frac{(\sec(\theta) - 1) \cos(\theta)}{\sin(\theta)} \]

This simplifies to:
\[ f(x) = \tan^{-1} \left( \frac{1 - \cos(\theta)}{\sin(\theta)} \right) \]

Using the half-angle identity, \( 1 - \cos(\theta) = 2\sin^2\left(\frac{\theta}{2}\right) \) and \( \sin(\theta) = 2\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right) \):
\[ f(x) = \tan^{-1} \left( \frac{2\sin^2\left(\frac{\theta}{2}\right)}{2\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right)} \right) = \tan^{-1} \left( \tan\left(\frac{\theta}{2}\right) \right) = \frac{\theta}{2} \]

Since \( \theta = \tan^{-1}(x) \):
\[ f(x) = \frac{1}{2} \tan^{-1}(x) \]

### Step 2: Differentiate \( f(x) \)

Now, we differentiate \( f(x) \):
\[ f'(x) = \frac{1}{2} \cdot \frac{1}{1+x^2} \]

### Step 3: Simplify \( g(x) \)

Next, we simplify \( g(x) \):
\[ g(x) = \tan^{-1} \left( \frac{2x \sqrt{1-x^2}}{1-2x^2} \right) \]

Let \( x = \sin(\theta) \). Then:
\[ g(x) = \tan^{-1} \left( \frac{2\sin(\theta)\cos(\theta)}{\cos^2(\theta)} \right) = \tan^{-1} \left( \tan(2\theta) \right) = 2\theta \]
Thus,
\[ g(x) = 2 \sin^{-1}(x) \]

### Step 4: Differentiate \( g(x) \)

Now we differentiate \( g(x) \):
\[ g'(x) = 2 \cdot \frac{1}{\sqrt{1-x^2}} \]

### Step 5: Find \( \frac{f'(x)}{g'(x)} \)

Now we find the derivative of \( f \) with respect to \( g \):
\[ \frac{f'(x)}{g'(x)} = \frac{\frac{1}{2(1+x^2)}}{2 \cdot \frac{1}{\sqrt{1-x^2}}} = \frac{\sqrt{1-x^2}}{4(1+x^2)} \]

### Final Answer

Thus, the derivative of \( f(x) \) with respect to \( g(x) \) is:
\[ \frac{\sqrt{1-x^2}}{4(1+x^2)} \]