1. Given function

$y=\ln\left(\frac{1-x^2}{1+x^2}\right)$

Using log properties,

$y=\ln(1-x^2)-\ln(1+x^2)$

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2. Find the first derivative

Differentiate term by term:

$y' = \frac{-2x}{1-x^2}-\frac{2x}{1+x^2}$

Take LCM and simplify:

$y'=-2x\left(\frac{1}{1-x^2}+\frac{1}{1+x^2}\right)$

$= -2x\left(\frac{(1+x^2)+(1-x^2)}{(1-x^2)(1+x^2)}\right)$

$= -2x\left(\frac{2}{1-x^4}\right)$

$\boxed{y'=-\frac{4x}{1-x^4}}$

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3. Find the second derivative

We differentiate

$y'=-\frac{4x}{1-x^4}$

Using quotient rule with

$u=-4x, \quad v=1-x^4$

so

$u'=-4, \quad v'=-4x^3$

Then

$y''=\frac{u'v-uv'}{v^2}$

$y''=\frac{-4(1-x^4)-(-4x)(-4x^3)}{(1-x^4)^2}$

$=\frac{-4+4x^4-16x^4}{(1-x^4)^2}$

$=\frac{-4-12x^4}{(1-x^4)^2}$

$\boxed{y''=-\frac{4(1+3x^4)}{(1-x^4)^2}}$

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4. Evaluate at $x=\frac12$

First,

$x=\frac12, \quad x^4=\frac{1}{16}$

Hence,

$1-x^4=1-\frac{1}{16}=\frac{15}{16}$

First derivative:

$y'=-\frac{4\cdot \frac12}{\frac{15}{16}}=-\frac{2}{\frac{15}{16}}=-\frac{32}{15}$

Second derivative:

$y''=-\frac{4\left(1+3\cdot \frac{1}{16}\right)}{\left(\frac{15}{16}\right)^2}$

$= -\frac{4\left(\frac{19}{16}\right)}{\frac{225}{256}}$

$= -\frac{19}{4}\cdot \frac{256}{225}$

$= -\frac{1216}{225}$

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5. Compute $225(y'-y'')$

$y'-y''=-\frac{32}{15}-\left(-\frac{1216}{225}\right)$

$= -\frac{480}{225}+\frac{1216}{225}$

$= \frac{736}{225}$

Therefore,

$225(y'-y'')=225\cdot \frac{736}{225}=\boxed{736}$

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6. Check options

• A: 732
• B: 736
• C: 742
• D: 746

So the correct option is:

$\boxed{\text{B: }736}$