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

Options

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

Answer

To find the derivative \(\frac{dy}{dx}\) of the function \(y = \left(1 + \frac{1}{x}\right)^x\), we can 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:

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

### Step 2: Apply the power rule of logarithms
Using the property of logarithms that states \(\ln(a^b) = b \ln a\), we can simplify the right side:

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

### Step 3: Differentiate both sides
Now we differentiate both sides with respect to \(x\). We will use the chain rule on the left side and the product rule on the right side:

\[ \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}\left(x \ln\left(1 + \frac{1}{x}\right)\right) \]

Using the product rule on the right side:

\[ \frac{d}{dx}\left(x \ln\left(1 + \frac{1}{x}\right)\right) = \ln\left(1 + \frac{1}{x}\right) + x \cdot \frac{d}{dx}\left(\ln\left(1 + \frac{1}{x}\right)\right) \]

### Step 4: Differentiate \(\ln\left(1 + \frac{1}{x}\right)\)
To differentiate \(\ln\left(1 + \frac{1}{x}\right)\), we use the chain rule:

\[ \frac{d}{dx}\left(\ln\left(1 + \frac{1}{x}\right)\right) = \frac{1}{1 + \frac{1}{x}} \cdot \frac{d}{dx}\left(1 + \frac{1}{x}\right) \]

The derivative of \(1 + \frac{1}{x}\) is \(-\frac{1}{x^2}\):

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

### Step 5: Substitute back into the derivative expression
Now substitute this back into the expression we derived in Step 3:

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

### Step 6: Solve for \(\frac{dy}{dx}\)
Multiply both sides by \(y\):

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

### Step 7: Substitute \(y\) back into the equation
Recall that \(y = \left(1 + \frac{1}{x}\right)^x\):

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

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

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