Let `y_1=tan^(-1)((sqrt(1+x^2)-1)/x)` and `y_2=tan^(-1)((2xsqrt(1-x^2))/(1-2x^2))` then `(dy_1)/(dy_2)=`

Options

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

Answer

To solve the problem, we need to find the derivative \(\frac{dy_1}{dy_2}\) where:

\[ y_1 = \tan^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right) \]
\[ y_2 = \tan^{-1}\left(\frac{2x\sqrt{1-x^2}}{1-2x^2}\right) \]

### Step 1: Differentiate \(y_1\)

We start by differentiating \(y_1\) with respect to \(x\):

\[ y_1 = \tan^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right) \]

Using the chain rule, the derivative of \(\tan^{-1}(u)\) is \(\frac{1}{1+u^2} \cdot \frac{du}{dx}\). Let \(u = \frac{\sqrt{1+x^2}-1}{x}\).

First, we find \(\frac{du}{dx}\):

\[ u = \frac{\sqrt{1+x^2}-1}{x} \]

Using the quotient rule:

\[ \frac{du}{dx} = \frac{x \cdot \frac{d}{dx}(\sqrt{1+x^2}-1) - (\sqrt{1+x^2}-1) \cdot \frac{d}{dx}(x)}{x^2} \]

Calculating \(\frac{d}{dx}(\sqrt{1+x^2})\):

\[ \frac{d}{dx}(\sqrt{1+x^2}) = \frac{x}{\sqrt{1+x^2}} \]

So,

\[ \frac{du}{dx} = \frac{x \cdot \frac{x}{\sqrt{1+x^2}} - \left(\sqrt{1+x^2}-1\right)}{x^2} \]

Simplifying this gives:

\[ \frac{du}{dx} = \frac{x^2/\sqrt{1+x^2} - \sqrt{1+x^2} + 1}{x^2} \]

Now, substituting \(u\) back into the derivative of \(y_1\):

\[ \frac{dy_1}{dx} = \frac{1}{1+u^2} \cdot \frac{du}{dx} \]

### Step 2: Differentiate \(y_2\)

Next, we differentiate \(y_2\):

\[ y_2 = \tan^{-1}\left(\frac{2x\sqrt{1-x^2}}{1-2x^2}\right) \]

Let \(v = \frac{2x\sqrt{1-x^2}}{1-2x^2}\).

Using the chain rule again:

\[ \frac{dy_2}{dx} = \frac{1}{1+v^2} \cdot \frac{dv}{dx} \]

Calculating \(\frac{dv}{dx}\) using the quotient rule:

\[ \frac{dv}{dx} = \frac{(1-2x^2)(2\sqrt{1-x^2} + 2x \cdot \frac{-x}{\sqrt{1-x^2}}) - 2x\sqrt{1-x^2}(-4x)}{(1-2x^2)^2} \]

### Step 3: Find \(\frac{dy_1}{dy_2}\)

Now, we can find \(\frac{dy_1}{dy_2}\):

\[ \frac{dy_1}{dy_2} = \frac{\frac{dy_1}{dx}}{\frac{dy_2}{dx}} \]

Substituting the derivatives we calculated earlier into this expression will give us the final result.

### Final Result

After performing the calculations and simplifications, we arrive at:

\[ \frac{dy_1}{dy_2} = \frac{\sqrt{1-x^2}}{4(1+x^2)} \]