Differentiate `x sin ^-1x` with respect to ` cos^-1 x ` .

Answer

To differentiate \( y = x \sin^{-1}(x) \) with respect to \( \cos^{-1}(x) \), we will follow these steps:

### Step 1: Define the functions
Let:
- \( y_1 = x \sin^{-1}(x) \)
- \( y_2 = \cos^{-1}(x) \)

### Step 2: Differentiate \( y_1 \) with respect to \( x \)
Using the product rule for differentiation, where \( u = x \) and \( v = \sin^{-1}(x) \):
\[ \frac{dy_1}{dx} = \frac{d}{dx}(x) \cdot \sin^{-1}(x) + x \cdot \frac{d}{dx}(\sin^{-1}(x)) \]
Calculating each derivative:
- \( \frac{d}{dx}(x) = 1 \)
- \( \frac{d}{dx}(\sin^{-1}(x)) = \frac{1}{\sqrt{1 - x^2}} \)

Thus, we have:
\[ \frac{dy_1}{dx} = \sin^{-1}(x) + x \cdot \frac{1}{\sqrt{1 - x^2}} \]

### Step 3: Differentiate \( y_2 \) with respect to \( x \)
The derivative of \( y_2 = \cos^{-1}(x) \) is:
\[ \frac{dy_2}{dx} = -\frac{1}{\sqrt{1 - x^2}} \]

### Step 4: Find \( \frac{dy_1}{dy_2} \)
Using the chain rule:
\[ \frac{dy_1}{dy_2} = \frac{\frac{dy_1}{dx}}{\frac{dy_2}{dx}} \]
Substituting the derivatives we found:
\[ \frac{dy_1}{dy_2} = \frac{\sin^{-1}(x) + \frac{x}{\sqrt{1 - x^2}}}{-\frac{1}{\sqrt{1 - x^2}}} \]

### Step 5: Simplify the expression
To simplify:
\[ \frac{dy_1}{dy_2} = -\left(\sin^{-1}(x) + \frac{x}{\sqrt{1 - x^2}}\right) \cdot \sqrt{1 - x^2} \]
Distributing the negative:
\[ \frac{dy_1}{dy_2} = -\sin^{-1}(x) \sqrt{1 - x^2} - x \]

### Final Answer
Thus, the derivative of \( x \sin^{-1}(x) \) with respect to \( \cos^{-1}(x) \) is:
\[ \frac{dy_1}{dy_2} = -\sin^{-1}(x) \sqrt{1 - x^2} - x \]