Find the derivative of the following function with respect to 'x'
`e^(ax).sin^-1bx`

Answer

To find the derivative of the function \( y = e^{ax} \sin^{-1}(bx) \) 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 \( u = e^{ax} \) and \( v = \sin^{-1}(bx) \).

2. **Differentiate \( u \)**:
To find \( \frac{du}{dx} \):
\[ \frac{du}{dx} = \frac{d}{dx}(e^{ax}) = ae^{ax} \]

3. **Differentiate \( v \)**:
To find \( \frac{dv}{dx} \):
Using the chain rule, we have:
\[ \frac{dv}{dx} = \frac{d}{dx}(\sin^{-1}(bx)) = \frac{1}{\sqrt{1 - (bx)^2}} \cdot \frac{d}{dx}(bx) = \frac{b}{\sqrt{1 - b^2x^2}} \]

4. **Apply the Product Rule**:
Now, substitute \( u \), \( v \), \( \frac{du}{dx} \), and \( \frac{dv}{dx} \) into the product rule:
\[ \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \]
\[ = e^{ax} \cdot \frac{b}{\sqrt{1 - b^2x^2}} + \sin^{-1}(bx) \cdot ae^{ax} \]

5. **Combine the Terms**:
Thus, the derivative is:
\[ \frac{dy}{dx} = \frac{be^{ax}}{\sqrt{1 - b^2x^2}} + ae^{ax} \sin^{-1}(bx) \]

### Final Answer:
\[ \frac{dy}{dx} = \frac{be^{ax}}{\sqrt{1 - b^2x^2}} + ae^{ax} \sin^{-1}(bx) \]