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

Options

A. ` (x+(1)/(x))^(x) ((x^(2)-1)/( x^(2)+1) +log (x+(1)/(x))+x^((x+( 1)/(x)) )( (x^(2) +1)/( x^(2))- (( x^(2)-1)/( x^(2)))log x) `
B. ` (x+(1)/(x))^(x) ((x^(2)-1)/( x^(2)+1) +log (x+(1)/(x))+x^((x+( 1)/(x)) )( (x^(2) +1)/( x^(2))+ (( x^(2)-1)/( x^(2)))log x) `
C. ` (x+(1)/(x))^(x) ((x^(2)-1)/( x^(2)+1) +log (x+(1)/(x))+x^((x+( 1)/(x)) )( (x^(2) +1)/( x)- (( x^(2)-1)/( x^(2)))log x) `
D. ` (x+(1)/(x))^(x) ((x^(2)-1)/( x^(2)+1) +log (x+(1)/(x))+x^((x+( 1)/(x)) )( (x^(2) +1)/( x)+ (( x^(2)-1)/( x^(2)))log x) `

Answer

To solve the problem, we will differentiate the function given by:

\[ y = \left( x + \frac{1}{x} \right)^x + x^{\left( x + \frac{1}{x} \right)} \]

We will break this down into two parts, \( u \) and \( v \):

1. Let \( u = \left( x + \frac{1}{x} \right)^x \)
2. Let \( v = x^{\left( x + \frac{1}{x} \right)} \)

Then, we can express \( y \) as:

\[ y = u + v \]

### Step 1: Differentiate \( u \)

To differentiate \( u \), we will use logarithmic differentiation.

Taking the natural logarithm of both sides:

\[ \ln u = x \ln \left( x + \frac{1}{x} \right) \]

Now, differentiate both sides with respect to \( x \):

Using the product rule on the right side:

\[ \frac{1}{u} \frac{du}{dx} = \ln \left( x + \frac{1}{x} \right) + x \cdot \frac{1}{x + \frac{1}{x}} \left( 1 - \frac{1}{x^2} \right) \]

Now, simplify:

\[ \frac{du}{dx} = u \left( \ln \left( x + \frac{1}{x} \right) + \frac{x \left( 1 - \frac{1}{x^2} \right)}{x + \frac{1}{x}} \right) \]

Substituting back for \( u \):

\[ \frac{du}{dx} = \left( x + \frac{1}{x} \right)^x \left( \ln \left( x + \frac{1}{x} \right) + \frac{x \left( 1 - \frac{1}{x^2} \right)}{x + \frac{1}{x}} \right) \]

### Step 2: Differentiate \( v \)

Now, we differentiate \( v \):

Taking the natural logarithm:

\[ \ln v = \left( x + \frac{1}{x} \right) \ln x \]

Differentiating both sides:

\[ \frac{1}{v} \frac{dv}{dx} = \left( 1 - \frac{1}{x^2} \right) \ln x + \left( x + \frac{1}{x} \right) \cdot \frac{1}{x} \]

Now, simplifying gives:

\[ \frac{dv}{dx} = v \left( \left( 1 - \frac{1}{x^2} \right) \ln x + \frac{x + \frac{1}{x}}{x} \right) \]

Substituting back for \( v \):

\[ \frac{dv}{dx} = x^{\left( x + \frac{1}{x} \right)} \left( \left( 1 - \frac{1}{x^2} \right) \ln x + \frac{x + \frac{1}{x}}{x} \right) \]

### Step 3: Combine the derivatives

Now, we can combine the derivatives to find \( \frac{dy}{dx} \):

\[ \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} \]

Substituting the expressions we derived for \( \frac{du}{dx} \) and \( \frac{dv}{dx} \):

\[ \frac{dy}{dx} = \left( x + \frac{1}{x} \right)^x \left( \ln \left( x + \frac{1}{x} \right) + \frac{x \left( 1 - \frac{1}{x^2} \right)}{x + \frac{1}{x}} \right) + x^{\left( x + \frac{1}{x} \right)} \left( \left( 1 - \frac{1}{x^2} \right) \ln x + \frac{x + \frac{1}{x}}{x} \right) \]

### Final Result

Thus, the derivative \( \frac{dy}{dx} \) is:

\[ \frac{dy}{dx} = \left( x + \frac{1}{x} \right)^x \left( \ln \left( x + \frac{1}{x} \right) + \frac{x \left( 1 - \frac{1}{x^2} \right)}{x + \frac{1}{x}} \right) + x^{\left( x + \frac{1}{x} \right)} \left( \left( 1 - \frac{1}{x^2} \right) \ln x + \frac{x + \frac{1}{x}}{x} \right) \]