Differentiate the following w.r.t. x :
`x^(cos^(-1)x)`

Answer

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

### Step 1: Take the natural logarithm of both sides
We start by taking the natural logarithm of both sides to simplify the differentiation process:
\[ \ln y = \ln(x^{\cos^{-1} x}) \]

### Step 2: Use the properties of logarithms
Using the property of logarithms that allows us to bring the exponent down, we rewrite the equation:
\[ \ln y = \cos^{-1} x \cdot \ln x \]

### Step 3: Differentiate both sides
Now, we differentiate both sides with respect to \( x \). We will use implicit differentiation on the left side and the product rule on the right side:
\[ \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(\cos^{-1} x) \cdot \ln x + \cos^{-1} x \cdot \frac{d}{dx}(\ln x) \]

### Step 4: Differentiate the components
We know that:
\[ \frac{d}{dx}(\cos^{-1} x) = -\frac{1}{\sqrt{1 - x^2}} \]
and
\[ \frac{d}{dx}(\ln x) = \frac{1}{x} \]
Substituting these derivatives into our equation gives:
\[ \frac{1}{y} \frac{dy}{dx} = -\frac{1}{\sqrt{1 - x^2}} \cdot \ln x + \cos^{-1} x \cdot \frac{1}{x} \]

### Step 5: Solve for \(\frac{dy}{dx}\)
Now, we can solve for \(\frac{dy}{dx}\) by multiplying both sides by \( y \):
\[ \frac{dy}{dx} = y \left( -\frac{\ln x}{\sqrt{1 - x^2}} + \frac{\cos^{-1} x}{x} \right) \]

### Step 6: Substitute back for \( y \)
Recall that \( y = x^{\cos^{-1} x} \), so we substitute back:
\[ \frac{dy}{dx} = x^{\cos^{-1} x} \left( -\frac{\ln x}{\sqrt{1 - x^2}} + \frac{\cos^{-1} x}{x} \right) \]

### Final Answer
Thus, the derivative of \( y = x^{\cos^{-1} x} \) with respect to \( x \) is:
\[ \frac{dy}{dx} = x^{\cos^{-1} x} \left( -\frac{\ln x}{\sqrt{1 - x^2}} + \frac{\cos^{-1} x}{x} \right) \]

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