Find the differential coefficient of thefollowing functions `x^(sin^-1 x)` w.r.t. `sin^-1 x`

Answer

To find the differential coefficient of the function \( u = x^{\sin^{-1} x} \) with respect to \( v = \sin^{-1} x \), we will follow these steps:

### Step 1: Define the Functions
Let:
- \( u = x^{\sin^{-1} x} \)
- \( v = \sin^{-1} x \)

### Step 2: Take the Natural Logarithm
Taking the natural logarithm of both sides:
\[ \ln u = \ln(x^{\sin^{-1} x}) \]
Using the property of logarithms \( \ln(m^n) = n \ln m \):
\[ \ln u = \sin^{-1} x \cdot \ln x \]

### Step 3: Differentiate Both Sides with Respect to \( x \)
Differentiating both sides with respect to \( x \):
\[ \frac{d}{dx}(\ln u) = \frac{d}{dx}(\sin^{-1} x \cdot \ln x) \]
Using the chain rule on the left side:
\[ \frac{1}{u} \frac{du}{dx} = \frac{d}{dx}(\sin^{-1} x) \cdot \ln x + \sin^{-1} x \cdot \frac{d}{dx}(\ln x) \]

### Step 4: Differentiate the Right Side
Now, we differentiate the right side:
- The derivative of \( \sin^{-1} x \) is \( \frac{1}{\sqrt{1 - x^2}} \)
- The derivative of \( \ln x \) is \( \frac{1}{x} \)

Thus, we have:
\[ \frac{1}{u} \frac{du}{dx} = \frac{1}{\sqrt{1 - x^2}} \cdot \ln x + \sin^{-1} x \cdot \frac{1}{x} \]

### Step 5: Solve for \( \frac{du}{dx} \)
Multiplying both sides by \( u \):
\[ \frac{du}{dx} = u \left( \frac{\ln x}{\sqrt{1 - x^2}} + \frac{\sin^{-1} x}{x} \right) \]
Substituting back \( u = x^{\sin^{-1} x} \):
\[ \frac{du}{dx} = x^{\sin^{-1} x} \left( \frac{\ln x}{\sqrt{1 - x^2}} + \frac{\sin^{-1} x}{x} \right) \]

### Step 6: Differentiate \( v \)
Now, we differentiate \( v \):
\[ \frac{dv}{dx} = \frac{d}{dx}(\sin^{-1} x) = \frac{1}{\sqrt{1 - x^2}} \]

### Step 7: Find \( \frac{du}{dv} \)
Using the formula \( \frac{du}{dv} = \frac{du/dx}{dv/dx} \):
\[ \frac{du}{dv} = \frac{x^{\sin^{-1} x} \left( \frac{\ln x}{\sqrt{1 - x^2}} + \frac{\sin^{-1} x}{x} \right)}{\frac{1}{\sqrt{1 - x^2}}} \]
This simplifies to:
\[ \frac{du}{dv} = x^{\sin^{-1} x} \left( \ln x + \frac{\sin^{-1} x}{x} \sqrt{1 - x^2} \right) \]

### Final Answer
Thus, the differential coefficient of \( u = x^{\sin^{-1} x} \) with respect to \( v = \sin^{-1} x \) is:
\[ \frac{du}{dv} = x^{\sin^{-1} x} \left( \ln x + \frac{\sin^{-1} x}{x} \sqrt{1 - x^2} \right) \]

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