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

Answer

To differentiate the function \( y = e^{\sqrt{1 - x^2}} \tan x \) with respect to \( x \), we will use the product rule of differentiation. The product rule states that if you have two functions \( u \) and \( v \), then the derivative of their product is given by:

\[ \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \]

### Step-by-Step Solution:

1. **Identify the functions**:
Let \( u = e^{\sqrt{1 - x^2}} \) and \( v = \tan x \).

2. **Differentiate \( u \)**:
To differentiate \( u \), we need to apply the chain rule:
\[ \frac{du}{dx} = e^{\sqrt{1 - x^2}} \cdot \frac{d}{dx}(\sqrt{1 - x^2}) \]
Now, differentiate \( \sqrt{1 - x^2} \):
\[ \frac{d}{dx}(\sqrt{1 - x^2}) = \frac{1}{2\sqrt{1 - x^2}} \cdot (-2x) = \frac{-x}{\sqrt{1 - x^2}} \]
Therefore,
\[ \frac{du}{dx} = e^{\sqrt{1 - x^2}} \cdot \frac{-x}{\sqrt{1 - x^2}} \]

3. **Differentiate \( v \)**:
The derivative of \( v = \tan x \) is:
\[ \frac{dv}{dx} = \sec^2 x \]

4. **Apply the product rule**:
Now, substituting \( u \), \( v \), \( \frac{du}{dx} \), and \( \frac{dv}{dx} \) into the product rule:
\[ \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \]
This gives:
\[ \frac{dy}{dx} = e^{\sqrt{1 - x^2}} \cdot \sec^2 x + \tan x \cdot \left( e^{\sqrt{1 - x^2}} \cdot \frac{-x}{\sqrt{1 - x^2}} \right) \]

5. **Simplify the expression**:
We can factor out \( e^{\sqrt{1 - x^2}} \):
\[ \frac{dy}{dx} = e^{\sqrt{1 - x^2}} \left( \sec^2 x - \frac{x \tan x}{\sqrt{1 - x^2}} \right) \]

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