Differentiate the following w.r.t. x :
`e^(cot^(-1)x^(2))`

Answer

To differentiate the function \( y = e^{\cot^{-1}(x^2)} \) with respect to \( x \), we can follow these steps:

### Step 1: Take the natural logarithm of both sides
We start by taking the natural logarithm of both sides to simplify the differentiation process.

\[ \ln y = \ln(e^{\cot^{-1}(x^2)}) \]

### Step 2: Use the property of logarithms
Using the property of logarithms, we can simplify the right-hand side:

\[ \ln y = \cot^{-1}(x^2) \cdot \ln e \]

Since \( \ln e = 1 \), we have:

\[ \ln y = \cot^{-1}(x^2) \]

### Step 3: Differentiate both sides with respect to \( x \)
Now we differentiate both sides with respect to \( x \). We use implicit differentiation on the left side and the chain rule on the right side.

\[ \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(\cot^{-1}(x^2)) \]

### Step 4: Differentiate \( \cot^{-1}(x^2) \)
To differentiate \( \cot^{-1}(x^2) \), we use the derivative formula:

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

where \( u = x^2 \). Therefore, \( \frac{du}{dx} = 2x \).

So, we have:

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

### Step 5: Substitute back into the equation
Now substituting this back into our differentiated equation:

\[ \frac{1}{y} \frac{dy}{dx} = -\frac{2x}{1 + x^4} \]

### Step 6: Solve for \( \frac{dy}{dx} \)
Multiplying both sides by \( y \):

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

### Step 7: Substitute \( y \) back in
Recall that \( y = e^{\cot^{-1}(x^2)} \):

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

### Final Result
Thus, the derivative of \( y = e^{\cot^{-1}(x^2)} \) with respect to \( x \) is:

\[ \frac{dy}{dx} = -\frac{2x e^{\cot^{-1}(x^2)}}{1 + x^4} \]