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

Answer

To solve the equation \( e^x \log y = \sin^{-1} x + \sin^{-1} y \) and find \(\frac{dy}{dx}\), we will differentiate both sides with respect to \(x\). Here’s a step-by-step solution:

### Step 1: Differentiate both sides
We start with the equation:
\[ e^x \log y = \sin^{-1} x + \sin^{-1} y \]
Now, we differentiate both sides 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
Using the product rule on the left side:
\[ \frac{d}{dx}(e^x) \cdot \log y + e^x \cdot \frac{d}{dx}(\log y) = \frac{d}{dx}(\sin^{-1} x) + \frac{d}{dx}(\sin^{-1} y) \]
This gives us:
\[ e^x \log y' + e^x \cdot \frac{1}{y} \frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}} + \frac{1}{\sqrt{1 - y^2}} \frac{dy}{dx} \]

### Step 3: Rearranging the equation
Rearranging the equation to isolate \(\frac{dy}{dx}\):
\[ e^x \cdot \frac{1}{y} \frac{dy}{dx} - \frac{1}{\sqrt{1 - y^2}} \frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}} - e^x \log y \]

### Step 4: Factor out \(\frac{dy}{dx}\)
Factoring out \(\frac{dy}{dx}\):
\[ \left( e^x \cdot \frac{1}{y} - \frac{1}{\sqrt{1 - y^2}} \right) \frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}} - e^x \log y \]

### Step 5: Solve for \(\frac{dy}{dx}\)
Now, we can solve for \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{\frac{1}{\sqrt{1 - x^2}} - e^x \log y}{e^x \cdot \frac{1}{y} - \frac{1}{\sqrt{1 - y^2}}} \]

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