Let `y =log_e(frac{1-x^2}{1+x^2})`, `-1 < x < 1`. Then at `x = 1/2` the value of `225(y' - y'')` is equal to

Options

A. 732
B. 746
C. 742
D. 736

Answer

To solve the problem, we need to find the value of \( 225(y' - y'') \) at \( x = \frac{1}{2} \) where \( y = \log_e\left(\frac{1 - x^2}{1 + x^2}\right) \).

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

Given:
\[ y = \log_e\left(\frac{1 - x^2}{1 + x^2}\right) \]

Using the properties of logarithms, we can rewrite \( y \):
\[ y = \log_e(1 - x^2) - \log_e(1 + x^2) \]

Now, we differentiate \( y \):
\[ y' = \frac{d}{dx} \left(\log_e(1 - x^2)\right) - \frac{d}{dx} \left(\log_e(1 + x^2)\right) \]

Using the derivative of \( \log_e(u) \) which is \( \frac{1}{u} \cdot \frac{du}{dx} \):
\[ y' = \frac{-2x}{1 - x^2} - \frac{2x}{1 + x^2} \]

### Step 2: Simplify \( y' \)

Now, we need to combine the two fractions:
\[ y' = \frac{-2x(1 + x^2) - 2x(1 - x^2)}{(1 - x^2)(1 + x^2)} \]
\[ = \frac{-2x - 2x^3 - 2x + 2x^3}{(1 - x^2)(1 + x^2)} \]
\[ = \frac{-4x}{1 - x^4} \]

### Step 3: Differentiate \( y' \) to find \( y'' \)

Now we differentiate \( y' \):
\[ y' = \frac{-4x}{1 - x^4} \]

Using the quotient rule:
\[ y'' = \frac{(1 - x^4)(-4) - (-4x)(-4x^3)}{(1 - x^4)^2} \]
\[ = \frac{-4(1 - x^4) - 16x^4}{(1 - x^4)^2} \]
\[ = \frac{-4 + 4x^4 - 16x^4}{(1 - x^4)^2} \]
\[ = \frac{-4 - 12x^4}{(1 - x^4)^2} \]

### Step 4: Evaluate \( y' \) and \( y'' \) at \( x = \frac{1}{2} \)

Now we substitute \( x = \frac{1}{2} \):
\[ y' \left(\frac{1}{2}\right) = \frac{-4 \cdot \frac{1}{2}}{1 - \left(\frac{1}{2}\right)^4} = \frac{-2}{1 - \frac{1}{16}} = \frac{-2}{\frac{15}{16}} = -\frac{32}{15} \]

\[ y'' \left(\frac{1}{2}\right) = \frac{-4 - 12 \cdot \left(\frac{1}{2}\right)^4}{(1 - \left(\frac{1}{2}\right)^4)^2} = \frac{-4 - 12 \cdot \frac{1}{16}}{(1 - \frac{1}{16})^2} = \frac{-4 - \frac{12}{16}}{\left(\frac{15}{16}\right)^2} \]
\[ = \frac{-4 - \frac{3}{4}}{\frac{225}{256}} = \frac{-\frac{16}{4} - \frac{3}{4}}{\frac{225}{256}} = \frac{-\frac{19}{4}}{\frac{225}{256}} = -\frac{19 \cdot 256}{4 \cdot 225} = -\frac{19 \cdot 64}{225} \]

### Step 5: Calculate \( 225(y' - y'') \)

Now we calculate \( y' - y'' \):
\[ y' - y'' = -\frac{32}{15} + \frac{19 \cdot 64}{225} \]

Finding a common denominator (225):
\[ = -\frac{32 \cdot 15}{225} + \frac{19 \cdot 64}{225} = \frac{-480 + 1216}{225} = \frac{736}{225} \]

Finally, we calculate:
\[ 225(y' - y'') = 225 \cdot \frac{736}{225} = 736 \]

### Final Answer
The value of \( 225(y' - y'') \) at \( x = \frac{1}{2} \) is \( \boxed{736} \).