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

Answer

To differentiate \( x \sin^{-1} x \) 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 \( x \)
To differentiate \( y_1 \), we will use the product rule. The product rule states that if you have two functions \( u \) and \( v \), then the derivative of their product is given by:
\[ \frac{d(uv)}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \]

Here, let:
- \( u = x \) and \( v = \sin^{-1} x \)

Now, we differentiate:
- \( \frac{du}{dx} = 1 \)
- \( \frac{dv}{dx} = \frac{1}{\sqrt{1 - x^2}} \)

Using the product rule:
\[ \frac{dy_1}{dx} = x \cdot \frac{1}{\sqrt{1 - x^2}} + \sin^{-1} x \cdot 1 \]
\[ \frac{dy_1}{dx} = \frac{x}{\sqrt{1 - x^2}} + \sin^{-1} x \]

### Step 3: Differentiate \( y_2 \) with respect to \( x \)
Now, we differentiate \( y_2 \):
\[ \frac{dy_2}{dx} = \frac{d}{dx}(\cos^{-1} x) = -\frac{1}{\sqrt{1 - x^2}} \]

### Step 4: Find \( \frac{dy_1}{dy_2} \)
To find the derivative of \( y_1 \) with respect to \( y_2 \), we use the chain rule:
\[ \frac{dy_1}{dy_2} = \frac{dy_1/dx}{dy_2/dx} \]

Substituting the derivatives we found:
\[ \frac{dy_1}{dy_2} = \frac{\frac{x}{\sqrt{1 - x^2}} + \sin^{-1} x}{-\frac{1}{\sqrt{1 - x^2}}} \]

### Step 5: Simplify the expression
Now, we simplify:
\[ \frac{dy_1}{dy_2} = -\left( \frac{x + \sin^{-1} x \sqrt{1 - x^2}}{1} \right) \]
Thus, we have:
\[ \frac{dy_1}{dy_2} = -\left( x + \sin^{-1} x \sqrt{1 - x^2} \right) \]

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