If `(x)/(x ^(2) - 3x + 2) ,` find `(d ^(2) y )/(dx ^(2)).`

Options

A. `(2)/(( x -1)^(3)) + (2)/((x-2) ^(3))`
B. `(1)/(( x -1)^(3)) + (1)/((x-1) ^(3))`
C. `(2)/(( x -1)^(3)) + (2)/((x+2) ^(3))`
D. `(-2)/(( x -1)^(3)) + (4)/((x-2) ^(3))`

Answer

To find the second derivative \(\frac{d^2y}{dx^2}\) of the function \(y = \frac{x}{x^2 - 3x + 2}\), we will follow these steps:

### Step 1: Simplify the function
First, we simplify the denominator:
\[ x^2 - 3x + 2 = (x - 1)(x - 2) \]
Thus, we can rewrite the function as:
\[ y = \frac{x}{(x - 1)(x - 2)} \]

### Step 2: Use partial fractions
Next, we express \(y\) in terms of partial fractions:
\[ \frac{x}{(x - 1)(x - 2)} = \frac{A}{x - 1} + \frac{B}{x - 2} \]
Multiplying through by the denominator \((x - 1)(x - 2)\) gives:
\[ x = A(x - 2) + B(x - 1) \]
Expanding the right-hand side:
\[ x = Ax - 2A + Bx - B = (A + B)x - (2A + B) \]
Setting coefficients equal gives us the system:
1. \(A + B = 1\)
2. \(-2A - B = 0\)

### Step 3: Solve the system of equations
From the first equation, we can express \(B\) in terms of \(A\):
\[ B = 1 - A \]
Substituting into the second equation:
\[ -2A - (1 - A) = 0 \implies -2A - 1 + A = 0 \implies -A - 1 = 0 \implies A = -1 \]
Substituting \(A\) back to find \(B\):
\[ B = 1 - (-1) = 2 \]
Thus, we have:
\[ \frac{x}{(x - 1)(x - 2)} = \frac{-1}{x - 1} + \frac{2}{x - 2} \]

### Step 4: Differentiate to find the first derivative
Now we differentiate \(y\):
\[ y = -\frac{1}{x - 1} + \frac{2}{x - 2} \]
Using the derivative formula \(\frac{d}{dx}\left(\frac{1}{u}\right) = -\frac{u'}{u^2}\):
\[ \frac{dy}{dx} = -\left(-\frac{1}{(x - 1)^2}\right) + 2\left(-\frac{1}{(x - 2)^2}\right) \]
Thus,
\[ \frac{dy}{dx} = \frac{1}{(x - 1)^2} - \frac{2}{(x - 2)^2} \]

### Step 5: Differentiate again to find the second derivative
Now we differentiate \(\frac{dy}{dx}\):
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{1}{(x - 1)^2}\right) - \frac{d}{dx}\left(\frac{2}{(x - 2)^2}\right) \]
Using the derivative formula again:
\[ \frac{d}{dx}\left(\frac{1}{(x - 1)^2}\right) = -\frac{-2}{(x - 1)^3} = \frac{2}{(x - 1)^3} \]
\[ \frac{d}{dx}\left(\frac{2}{(x - 2)^2}\right) = -\frac{2 \cdot -2}{(x - 2)^3} = \frac{4}{(x - 2)^3} \]
Thus,
\[ \frac{d^2y}{dx^2} = \frac{2}{(x - 1)^3} - \frac{4}{(x - 2)^3} \]

### Final Answer
The second derivative is:
\[ \frac{d^2y}{dx^2} = \frac{2}{(x - 1)^3} - \frac{4}{(x - 2)^3} \]