Differentiate the following w.r.t. x :
`e^(sin^(2))(2tan^(-1)sqrt((1-x)/(1+x)))`

Answer

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

### Step 1: Identify the outer and inner functions
The function can be seen as a composition of functions. The outer function is \( e^u \) where \( u = \sin^2(2\tan^{-1}(\sqrt{\frac{1-x}{1+x}})) \).

### Step 2: Differentiate the outer function
Using the chain rule, the derivative of \( e^u \) with respect to \( x \) is:
\[ \frac{dy}{dx} = e^u \cdot \frac{du}{dx} \]

### Step 3: Differentiate the inner function \( u \)
Now we need to differentiate \( u = \sin^2(2\tan^{-1}(\sqrt{\frac{1-x}{1+x}})) \). We will use the chain rule again:
\[ \frac{du}{dx} = 2\sin(2\tan^{-1}(\sqrt{\frac{1-x}{1+x}})) \cdot \cos(2\tan^{-1}(\sqrt{\frac{1-x}{1+x}})) \cdot \frac{d}{dx}(2\tan^{-1}(\sqrt{\frac{1-x}{1+x}})) \]

### Step 4: Differentiate \( 2\tan^{-1}(\sqrt{\frac{1-x}{1+x}}) \)
Let \( v = \tan^{-1}(\sqrt{\frac{1-x}{1+x}}) \). The derivative of \( \tan^{-1}(z) \) is \( \frac{1}{1+z^2} \cdot \frac{dz}{dx} \). We need to find \( \frac{dz}{dx} \):
\[ z = \sqrt{\frac{1-x}{1+x}} \]
Using the quotient rule:
\[ \frac{dz}{dx} = \frac{(1+x)(-\frac{1}{2\sqrt{(1-x)(1+x)}})(-1) - (1-x)(\frac{1}{2\sqrt{(1-x)(1+x)}})(1)}{(1+x)^2} \]
This simplifies to:
\[ \frac{dz}{dx} = \frac{1}{2\sqrt{(1-x)(1+x)}(1+x)^2} \]

### Step 5: Combine the derivatives
Now we can substitute back into our expression for \( \frac{du}{dx} \):
\[ \frac{du}{dx} = 2\sin(2v) \cdot \cos(2v) \cdot 2 \cdot \frac{1}{1+\left(\sqrt{\frac{1-x}{1+x}}\right)^2} \cdot \frac{dz}{dx} \]
Where \( \cos(2v) = 1 - 2\sin^2(v) \).

### Step 6: Substitute and simplify
Finally, we substitute back into our expression for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = e^{\sin^2(2\tan^{-1}(\sqrt{\frac{1-x}{1+x}}))} \cdot \frac{du}{dx} \]

### Final Result
The final expression for the derivative \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = e^{\sin^2(2\tan^{-1}(\sqrt{\frac{1-x}{1+x}}))} \cdot \text{(expression from step 5)} \]