Find `(dy)/(dx)`, when:
`y=x^((cos^(-1)x))`

Answer

To find \(\frac{dy}{dx}\) for the function \(y = x^{\cos^{-1}(x)}\), we will use logarithmic differentiation. Here’s a step-by-step solution:

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

### Step 2: Simplify using logarithmic properties
Using the property of logarithms that states \(\ln(a^b) = b \ln(a)\), we can simplify the right-hand side:
\[ \ln(y) = \cos^{-1}(x) \cdot \ln(x) \]

### Step 3: Differentiate both sides with respect to \(x\)
Now, we differentiate both sides with respect to \(x\). We will use the product rule on the right side:
\[ \frac{d}{dx}(\ln(y)) = \frac{d}{dx}(\cos^{-1}(x) \cdot \ln(x)) \]
Using the chain rule on the left 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 \(\cos^{-1}(x)\) and \(\ln(x)\)
We know:
\[ \frac{d}{dx}(\cos^{-1}(x)) = -\frac{1}{\sqrt{1-x^2}} \]
and
\[ \frac{d}{dx}(\ln(x)) = \frac{1}{x} \]
Substituting these derivatives into our equation gives:
\[ \frac{1}{y} \frac{dy}{dx} = -\frac{1}{\sqrt{1-x^2}} \ln(x) + \cos^{-1}(x) \cdot \frac{1}{x} \]

### Step 5: Solve for \(\frac{dy}{dx}\)
Now, we can solve for \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = y \left(-\frac{1}{\sqrt{1-x^2}} \ln(x) + \frac{\cos^{-1}(x)}{x}\right) \]
Substituting back the value of \(y\):
\[ \frac{dy}{dx} = x^{\cos^{-1}(x)} \left(-\frac{1}{\sqrt{1-x^2}} \ln(x) + \frac{\cos^{-1}(x)}{x}\right) \]

### Final Answer
Thus, the derivative \(\frac{dy}{dx}\) 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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