The derivative of `sin^(-1)((2x)/(1+x^(2)))` with respect to `cos^(-1)((1-x^(2))/(1+x^(2)))` is

Options

A. `-1`
B. 1
C. 2
D. `-2`

Explanation

To find the derivative of \( \sin^{-1}\left(\frac{2x}{1+x^2}\right) \) with respect to \( \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) \), we can use the chain rule. Let's denote:

- \( u = \sin^{-1}\left(\frac{2x}{1+x^2}\right) \)
- \( v = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) \)

We want to find \( \frac{du}{dv} \).

### Step 1: Find \( \frac{du}{dx} \)

Using the derivative of the inverse sine function, we have:

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

Now, we need to compute \( \frac{d}{dx}\left(\frac{2x}{1+x^2}\right) \) using the quotient rule:

\[ \frac{d}{dx}\left(\frac{2x}{1+x^2}\right) = \frac{(1+x^2)(2) - 2x(2x)}{(1+x^2)^2} = \frac{2 + 2x^2 - 4x^2}{(1+x^2)^2} = \frac{2 - 2x^2}{(1+x^2)^2} = \frac{2(1-x^2)}{(1+x^2)^2} \]

Now substituting this back into the expression for \( \frac{du}{dx} \):

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

### Step 2: Find \( \frac{dv}{dx} \)

Using the derivative of the inverse cosine function, we have:

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

Now, we compute \( \frac{d}{dx}\left(\frac{1-x^2}{1+x^2}\right) \) using the quotient rule:

\[ \frac{d}{dx}\left(\frac{1-x^2}{1+x^2}\right) = \frac{(1+x^2)(-2x) - (1-x^2)(2x)}{(1+x^2)^2} = \frac{-2x(1+x^2) - 2x(1-x^2)}{(1+x^2)^2} = \frac{-2x(1+x^2 + 1 - x^2)}{(1+x^2)^2} = \frac{-4x}{(1+x^2)^2} \]

Now substituting this back into the expression for \( \frac{dv}{dx} \):

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

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

Using the chain rule, we have:

\[ \frac{du}{dv} = \frac{du/dx}{dv/dx} \]

Substituting the expressions we derived:

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

### Step 4: Simplify the expression

The \( (1+x^2)^2 \) terms cancel out:

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

### Final Answer

Thus, the derivative of \( \sin^{-1}\left(\frac{2x}{1+x^2}\right) \) with respect to \( \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) \) is:

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