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 two functions \( u(x) \) and \( v(x) \), then the derivative of their product is given by:

\[ \frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x) \]

### Step-by-Step Solution:

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

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

3. **Apply the product rule**:
Using the product rule:
\[ \frac{dy}{dx} = u'v + uv' \]
Substituting the values we found:
\[ \frac{dy}{dx} = e^x \tan^{-1}(x) + e^x \cdot \frac{1}{1 + x^2} \]

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

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