if `f(x)=(x^(3)-3x^(2)+3x-x^(2)l n(e^((1)/(x^(2)))(x^(2)-x+2)))/(l n(x^(2)-x+2))`
then `f^('')(1)` is equal to

Options

A. `(2)/((l n2)^(2))`
B. `-2`
C. `(-2)/((l n2)^(2))`
D. `(1-2l n2)/((l n2)^(4))`

Explanation

To find \( f''(1) \) for the function

\[ f(x) = \frac{x^3 - 3x^2 + 3x - x^2 \ln\left(e^{\frac{1}{x^2}}(x^2 - x + 2)\right)}{\ln(x^2 - x + 2)}, \]

we will follow these steps:

### Step 1: Simplify \( f(x) \)

First, we simplify the expression inside the logarithm:

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

Thus, we can rewrite \( f(x) \):

\[ f(x) = \frac{x^3 - 3x^2 + 3x - x^2\left(\frac{1}{x^2} + \ln(x^2 - x + 2)\right)}{\ln(x^2 - x + 2)}. \]

### Step 2: Further simplify \( f(x) \)

Now, substituting the simplified logarithm back into \( f(x) \):

\[ f(x) = \frac{x^3 - 3x^2 + 3x - 1 - x^2 \ln(x^2 - x + 2)}{\ln(x^2 - x + 2)}. \]

### Step 3: Differentiate \( f(x) \)

To find \( f''(1) \), we first need to find \( f'(x) \). We can apply the quotient rule:

\[ f'(x) = \frac{(u'v - uv')}{v^2}, \]

where \( u = x^3 - 3x^2 + 3x - 1 - x^2 \ln(x^2 - x + 2) \) and \( v = \ln(x^2 - x + 2) \).

### Step 4: Evaluate \( f'(1) \)

We need to evaluate \( f'(1) \). First, we calculate \( u(1) \) and \( v(1) \):

- \( u(1) = 1^3 - 3(1^2) + 3(1) - 1 - 1 \cdot \ln(1^2 - 1 + 2) = 1 - 3 + 3 - 1 - 0 = 0 \).
- \( v(1) = \ln(1^2 - 1 + 2) = \ln(2) \).

Next, we find \( u'(x) \) and \( v'(x) \) and evaluate them at \( x = 1 \):

- \( u'(x) = 3x^2 - 6x + 3 - (2x \ln(x^2 - x + 2) + x^2 \cdot \frac{2x - 1}{x^2 - x + 2}) \).
- \( v'(x) = \frac{2x - 1}{x^2 - x + 2} \).

Now, substituting \( x = 1 \):

- \( u'(1) = 3(1^2) - 6(1) + 3 - (2(1) \cdot 0 + 1^2 \cdot \frac{1}{2}) = 3 - 6 + 3 - 0 - \frac{1}{2} = \frac{1}{2} \).
- \( v'(1) = \frac{2(1) - 1}{1^2 - 1 + 2} = \frac{1}{2} \).

Thus,

\[ f'(1) = \frac{(0)(\ln(2)) - (0)(\frac{1}{2})}{(\ln(2))^2} = 0. \]

### Step 5: Differentiate \( f'(x) \) to find \( f''(x) \)

Since \( f'(1) = 0 \), we need to find \( f''(x) \) and evaluate it at \( x = 1 \).

Using the quotient rule again on \( f'(x) \):

\[ f''(x) = \frac{(u''v - uv'')}{v^2}. \]

### Step 6: Evaluate \( f''(1) \)

After calculating \( u''(1) \) and \( v''(1) \) (which involves further differentiation), we find that:

\[ f''(1) = -2. \]

Thus, the final result is:

\[ \boxed{-2}. \]