If ` y=x^(tan ^(-1)x)` ,then `(dy)/(dx)=`

Options

A. ` y(( tan ^(-1) x)/( x ) +(logx )/( 1+x^(2)))`
B. ` y(( tan ^(-1) x)/( x ) -(logx )/( 1+x^(2)))`
C. `( y)/(2) (( tan ^(-1) x)/( x ) +(logx )/( 1+x^(2)))`
D. `( y)/(2) (( tan ^(-1) x)/( x ) -(logx )/( 1+x^(2)))`

Answer

To find the derivative of the function \( y = x^{\tan^{-1}(x)} \), we will use logarithmic differentiation. Here’s the 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:
\[ \ln(y) = \ln(x^{\tan^{-1}(x)}) \]

### Step 2: Apply the logarithmic power rule
Using the property of logarithms that states \( \ln(a^b) = b \cdot \ln(a) \), we can rewrite the right side:
\[ \ln(y) = \tan^{-1}(x) \cdot \ln(x) \]

### Step 3: Differentiate both sides with respect to \( x \)
Now we differentiate both sides. For the left side, we use implicit differentiation:
\[ \frac{1}{y} \cdot \frac{dy}{dx} = \frac{d}{dx}(\tan^{-1}(x) \cdot \ln(x)) \]

### Step 4: Apply the product rule on the right side
To differentiate the right side, we apply the product rule:
\[ \frac{d}{dx}(\tan^{-1}(x) \cdot \ln(x)) = \tan^{-1}(x) \cdot \frac{d}{dx}(\ln(x)) + \ln(x) \cdot \frac{d}{dx}(\tan^{-1}(x)) \]
Calculating the derivatives:
- The derivative of \( \ln(x) \) is \( \frac{1}{x} \).
- The derivative of \( \tan^{-1}(x) \) is \( \frac{1}{1+x^2} \).

So we have:
\[ \frac{d}{dx}(\tan^{-1}(x) \cdot \ln(x)) = \tan^{-1}(x) \cdot \frac{1}{x} + \ln(x) \cdot \frac{1}{1+x^2} \]

### Step 5: Substitute back into the equation
Substituting back, we get:
\[ \frac{1}{y} \cdot \frac{dy}{dx} = \tan^{-1}(x) \cdot \frac{1}{x} + \ln(x) \cdot \frac{1}{1+x^2} \]

### Step 6: Solve for \( \frac{dy}{dx} \)
Now, we multiply both sides by \( y \):
\[ \frac{dy}{dx} = y \left( \tan^{-1}(x) \cdot \frac{1}{x} + \ln(x) \cdot \frac{1}{1+x^2} \right) \]

### Step 7: Substitute \( y \) back into the equation
Recall that \( y = x^{\tan^{-1}(x)} \):
\[ \frac{dy}{dx} = x^{\tan^{-1}(x)} \left( \tan^{-1}(x) \cdot \frac{1}{x} + \ln(x) \cdot \frac{1}{1+x^2} \right) \]

### Final Result
Thus, the derivative \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = x^{\tan^{-1}(x)} \left( \frac{\tan^{-1}(x)}{x} + \frac{\ln(x)}{1+x^2} \right) \]