Differentiate the following w.r.t. x :
`cos^(-1)((2x)/(1+x^(2))),-1ltxlt1`

Answer

To differentiate the function \( y = \cos^{-1}\left(\frac{2x}{1+x^2}\right) \) with respect to \( x \), we will follow these steps:

### Step 1: Identify the function and apply the chain rule
Let \( u = \frac{2x}{1+x^2} \). Then, we can express \( y \) as:
\[ y = \cos^{-1}(u) \]
To find \(\frac{dy}{dx}\), we will use the chain rule:
\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \]

### Step 2: Differentiate \( y \) with respect to \( u \)
The derivative of \( \cos^{-1}(u) \) is:
\[ \frac{dy}{du} = -\frac{1}{\sqrt{1-u^2}} \]

### Step 3: Differentiate \( u \) with respect to \( x \)
Now we need to differentiate \( u = \frac{2x}{1+x^2} \). We will use the quotient rule:
\[ \frac{du}{dx} = \frac{(1+x^2)(2) - (2x)(2x)}{(1+x^2)^2} \]
Simplifying this:
\[ \frac{du}{dx} = \frac{2(1+x^2) - 4x^2}{(1+x^2)^2} = \frac{2 - 2x^2}{(1+x^2)^2} = \frac{2(1-x^2)}{(1+x^2)^2} \]

### Step 4: Substitute \( u \) back into the derivative
Now we can substitute \( u \) back into our expression for \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = -\frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx} \]
Substituting \( u = \frac{2x}{1+x^2} \):
\[ 1 - u^2 = 1 - \left(\frac{2x}{1+x^2}\right)^2 = 1 - \frac{4x^2}{(1+x^2)^2} \]
Finding a common denominator:
\[ 1 - u^2 = \frac{(1+x^2)^2 - 4x^2}{(1+x^2)^2} = \frac{1 + 2x^2 + x^4 - 4x^2}{(1+x^2)^2} = \frac{1 - 2x^2 + x^4}{(1+x^2)^2} = \frac{(1-x^2)^2}{(1+x^2)^2} \]
Thus,
\[ \sqrt{1 - u^2} = \frac{1-x^2}{1+x^2} \]

### Step 5: Combine everything to find \(\frac{dy}{dx}\)
Now substituting back:
\[ \frac{dy}{dx} = -\frac{1}{\frac{1-x^2}{1+x^2}} \cdot \frac{2(1-x^2)}{(1+x^2)^2} \]
This simplifies to:
\[ \frac{dy}{dx} = -\frac{2(1-x^2)}{(1+x^2)^2} \cdot \frac{1+x^2}{1-x^2} = -\frac{2}{1+x^2} \]

### Final Answer
Thus, the derivative of \( y = \cos^{-1}\left(\frac{2x}{1+x^2}\right) \) with respect to \( x \) is:
\[ \frac{dy}{dx} = -\frac{2}{1+x^2} \]