If `y=e^x tan^(-1)x`, then dy/dx is:

Options

A. `e^x[tan^(-1)x+1/(1+x^2)]`
B. `e^x[tan^(-1)x+1/(1-x^2)]`
C. `e^x[tan^(-1)x+1/sqrt(1+x^2)]`
D. `e^x[tan^(-1)x-1/(1+x^2)]`

Answer

To find the derivative of the function \( y = e^x \tan^{-1}(x) \), we will use the product rule of differentiation. The product rule states that if you have a function that is the product of two functions \( u \) and \( v \), then the derivative \( \frac{dy}{dx} \) is given by:

\[ \frac{dy}{dx} = u'v + uv' \]

### Step-by-step Solution:

1. **Identify the functions**:
- Let \( u = e^x \)
- Let \( v = \tan^{-1}(x) \)

2. **Differentiate \( u \) and \( v \)**:
- The derivative of \( u \) with respect to \( x \) is:
\[ u' = \frac{d}{dx}(e^x) = e^x \]
- The derivative of \( v \) with respect to \( x \) is:
\[ v' = \frac{d}{dx}(\tan^{-1}(x)) = \frac{1}{1 + x^2} \]

3. **Apply the product rule**:
- Now, using the product rule:
\[ \frac{dy}{dx} = u'v + uv' \]
- Substitute \( u \), \( u' \), \( v \), and \( v' \):
\[ \frac{dy}{dx} = e^x \tan^{-1}(x) + e^x \cdot \frac{1}{1 + x^2} \]

4. **Factor out \( e^x \)**:
- We can factor out \( e^x \) from both terms:
\[ \frac{dy}{dx} = e^x \left( \tan^{-1}(x) + \frac{1}{1 + x^2} \right) \]

### Final Answer:
Thus, the derivative \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = e^x \left( \tan^{-1}(x) + \frac{1}{1 + x^2} \right) \]