`e^xlogy=sin^(-1)x+sin^(-1)y`

Answer

To differentiate the given expression \( e^x \log y = \sin^{-1} x + \sin^{-1} y \) with respect to \( x \), we will follow these steps:

### Step 1: Differentiate both sides
We start by differentiating both sides of the equation with respect to \( x \):
\[ \frac{d}{dx}(e^x \log y) = \frac{d}{dx}(\sin^{-1} x + \sin^{-1} y) \]

### Step 2: Apply the product rule on the left side
The left side involves a product of two functions \( e^x \) and \( \log y \). We apply the product rule:
\[ \frac{d}{dx}(e^x \log y) = e^x \frac{d}{dx}(\log y) + \log y \frac{d}{dx}(e^x) \]
The derivative of \( \log y \) with respect to \( x \) is \( \frac{1}{y} \frac{dy}{dx} \) (using the chain rule), and the derivative of \( e^x \) is \( e^x \):
\[ = e^x \left(\frac{1}{y} \frac{dy}{dx}\right) + \log y \cdot e^x \]

### Step 3: Differentiate the right side
Now, we differentiate the right side:
\[ \frac{d}{dx}(\sin^{-1} x + \sin^{-1} y) = \frac{1}{\sqrt{1 - x^2}} + \frac{1}{\sqrt{1 - y^2}} \frac{dy}{dx} \]

### Step 4: Set the derivatives equal
Now we set the derivatives from both sides equal to each other:
\[ e^x \left(\frac{1}{y} \frac{dy}{dx}\right) + \log y \cdot e^x = \frac{1}{\sqrt{1 - x^2}} + \frac{1}{\sqrt{1 - y^2}} \frac{dy}{dx} \]

### Step 5: Collect terms involving \( \frac{dy}{dx} \)
Rearranging the equation to isolate \( \frac{dy}{dx} \):
\[ e^x \frac{1}{y} \frac{dy}{dx} - \frac{1}{\sqrt{1 - y^2}} \frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}} - \log y \cdot e^x \]

### Step 6: Factor out \( \frac{dy}{dx} \)
Factoring out \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} \left(e^x \frac{1}{y} - \frac{1}{\sqrt{1 - y^2}}\right) = \frac{1}{\sqrt{1 - x^2}} - \log y \cdot e^x \]

### Step 7: Solve for \( \frac{dy}{dx} \)
Now, we can solve for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{\frac{1}{\sqrt{1 - x^2}} - \log y \cdot e^x}{e^x \frac{1}{y} - \frac{1}{\sqrt{1 - y^2}}} \]

### Final Result
Thus, the derivative \( \frac{dy}{dx} \) is given by:
\[ \frac{dy}{dx} = \frac{y \left(\frac{1}{\sqrt{1 - x^2}} - \log y \cdot e^x\right)}{e^x \sqrt{1 - y^2} - y} \]