If `y=(x)/(x+1)+(x+1)/(x)," then "(d^(2)y)/(dx^(2))` at x=1 is equal to

Options

A. `(7)/(4)`
B. `(7)/(8)`
C. `(1)/(4)`
D. `(-7)/(8)`

Answer

To find \(\frac{d^2y}{dx^2}\) at \(x=1\) for the function \(y = \frac{x}{x+1} + \frac{x+1}{x}\), we will follow these steps:

### Step 1: Simplify the function
First, we simplify the expression for \(y\):
\[ y = \frac{x}{x+1} + \frac{x+1}{x} \]
We can rewrite the second term:
\[ y = \frac{x}{x+1} + \left(1 + \frac{1}{x}\right) = \frac{x}{x+1} + 1 + \frac{1}{x} \]
Combining these, we can express \(y\) as:
\[ y = 1 + \frac{x}{x+1} + \frac{1}{x} \]

### Step 2: Differentiate \(y\) to find \(\frac{dy}{dx}\)
Now we differentiate \(y\):
\[ \frac{dy}{dx} = \frac{d}{dx}\left(\frac{x}{x+1}\right) + \frac{d}{dx}\left(1\right) + \frac{d}{dx}\left(\frac{1}{x}\right) \]
Using the quotient rule for \(\frac{x}{x+1}\):
\[ \frac{d}{dx}\left(\frac{x}{x+1}\right) = \frac{(x+1)(1) - x(1)}{(x+1)^2} = \frac{1}{(x+1)^2} \]
And for \(\frac{1}{x}\):
\[ \frac{d}{dx}\left(\frac{1}{x}\right) = -\frac{1}{x^2} \]
Thus:
\[ \frac{dy}{dx} = \frac{1}{(x+1)^2} - \frac{1}{x^2} \]

### Step 3: Differentiate again to find \(\frac{d^2y}{dx^2}\)
Now we differentiate \(\frac{dy}{dx}\):
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{1}{(x+1)^2}\right) - \frac{d}{dx}\left(\frac{1}{x^2}\right) \]
Using the chain rule:
\[ \frac{d}{dx}\left(\frac{1}{(x+1)^2}\right) = -\frac{2}{(x+1)^3} \]
And:
\[ \frac{d}{dx}\left(\frac{1}{x^2}\right) = -\frac{2}{x^3} \]
Thus:
\[ \frac{d^2y}{dx^2} = -\frac{2}{(x+1)^3} + \frac{2}{x^3} \]

### Step 4: Evaluate \(\frac{d^2y}{dx^2}\) at \(x=1\)
Now we substitute \(x=1\):
\[ \frac{d^2y}{dx^2}\bigg|_{x=1} = -\frac{2}{(1+1)^3} + \frac{2}{1^3} \]
Calculating this:
\[ = -\frac{2}{2^3} + 2 = -\frac{2}{8} + 2 = -\frac{1}{4} + 2 = 2 - \frac{1}{4} = \frac{8}{4} - \frac{1}{4} = \frac{7}{4} \]

### Final Answer
Thus, \(\frac{d^2y}{dx^2}\) at \(x=1\) is:
\[ \frac{7}{4} \]