The derivative of `tan^(-1)((sqrt(1+x^2)-1)/x)` with respect to `tan^(-1)((2xsqrt(1-x^2))/(1-2x^2))` at `x=0` is

Options

A. `1/8`
B. `1/4`
C. `1/2`
D. `1`

Answer

To find the derivative of \( y = \tan^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right) \) with respect to \( z = \tan^{-1}\left(\frac{2x\sqrt{1-x^2}}{1-2x^2}\right) \) at \( x = 0 \), we will follow these steps:

### Step 1: Simplify \( y \)
Let \( x = \tan(\theta) \). Then, we have:
\[ \sqrt{1+x^2} = \sqrt{1+\tan^2(\theta)} = \sec(\theta) \]
Thus,
\[ y = \tan^{-1}\left(\frac{\sec(\theta) - 1}{\tan(\theta)}\right) = \tan^{-1}\left(\frac{\sec(\theta) - 1}{\tan(\theta)}\right) \]
Using the identity \( \sec(\theta) - 1 = \frac{\tan^2(\theta/2)}{\cos^2(\theta/2)} \), we can rewrite:
\[ y = \tan^{-1}\left(\frac{\tan^2(\theta/2)}{\sin(\theta)}\right) \]
This simplifies to:
\[ y = \frac{\theta}{2} \]
where \( \theta = \tan^{-1}(x) \).

### Step 2: Simplify \( z \)
Let \( x = \sin(y) \). Then:
\[ z = \tan^{-1}\left(\frac{2\sin(y)\cos(y)}{\cos(2y)}\right) = \tan^{-1}(\sin(2y)) \]
This can be simplified to:
\[ z = 2y \]

### Step 3: Differentiate \( y \) with respect to \( z \)
Using the chain rule:
\[ \frac{dy}{dz} = \frac{dy/dx}{dz/dx} \]
We need to find \( \frac{dy}{dx} \) and \( \frac{dz}{dx} \).

### Step 4: Calculate \( \frac{dy}{dx} \)
From \( y = \frac{1}{2}\tan^{-1}(x) \):
\[ \frac{dy}{dx} = \frac{1}{2} \cdot \frac{1}{1+x^2} \]

### Step 5: Calculate \( \frac{dz}{dx} \)
From \( z = 2 \sin^{-1}(x) \):
\[ \frac{dz}{dx} = 2 \cdot \frac{1}{\sqrt{1-x^2}} \]

### Step 6: Combine results
Now we have:
\[ \frac{dy}{dz} = \frac{\frac{1}{2(1+x^2)}}{2 \cdot \frac{1}{\sqrt{1-x^2}}} = \frac{1}{4} \cdot \frac{\sqrt{1-x^2}}{1+x^2} \]

### Step 7: Evaluate at \( x = 0 \)
Substituting \( x = 0 \):
\[ \frac{dy}{dz} \bigg|_{x=0} = \frac{1}{4} \cdot \frac{\sqrt{1-0^2}}{1+0^2} = \frac{1}{4} \]

### Final Answer
Thus, the derivative of \( y \) with respect to \( z \) at \( x = 0 \) is:
\[ \frac{1}{4} \]