Let y = _ e ( 1 - x ^ 2 1 + x ^ 2 ), - 1 < x < 1 . Then at x =...

Let, $f\left(x\right) = \log \left(\dfrac{1 - {x}^{2}}{1 + {x}^{2}}\right)$

$⇒ f\left(x\right) = \log \left(1 - {x}^{2}\right) - \log \left(1 + {x}^{2}\right)$

$⇒ {f}^{'}\left(x\right) = \dfrac{- 2x}{\left(1 - {x}^{2}\right)} - \dfrac{2x}{\left(1 + {x}^{2}\right)}$

$⇒ {f}^{'}\left(x\right) = \left(- 2x\right)\left[\dfrac{1 + {x}^{2} + 1 - {x}^{2}}{1 - {x}^{4}}\right]$

$⇒ {f}^{'}\left(x\right) = \left(\dfrac{4x}{{x}^{4} - 1}\right)$

$⇒ {f}^{"}\left(x\right) = \dfrac{\left({x}^{4} - 1\right)\left(4\right) - \left(4x\right)\left(4{x}^{3}\right)}{{\left({x}^{4} - 1\right)}^{2}}$

$⇒ {f}^{"}\left(x\right) = \dfrac{4\left(- 3{x}^{4} - 1\right)}{{\left({x}^{4} - 1\right)}^{2}}$

$⇒ 225\left[{f}^{'}\left(x\right) - {f}^{"}\left(x\right)\right] = 225\left[\left(\dfrac{4x}{{x}^{4} - 1}\right) - \dfrac{4\left(- 3{x}^{4} - 1\right)}{{\left({x}^{4} - 1\right)}^{2}}\right]$

Putting, $x = \dfrac{1}{2}$

$⇒ 225\left[{f}^{'}\left(\dfrac{1}{2}\right) - {f}^{"}\left(\dfrac{1}{2}\right)\right] = 225\left[\left(\dfrac{2}{- \dfrac{15}{16}}\right) - \dfrac{4\left(- \dfrac{3}{16} - 1\right)}{{\left(\dfrac{1}{16} - 1\right)}^{2}}\right]$

$⇒ 225\left[{f}^{'}\left(\dfrac{1}{2}\right) - {f}^{"}\left(\dfrac{1}{2}\right)\right] = 225\left[\left(\dfrac{- 32}{15}\right) - \dfrac{4\left(\dfrac{- 19}{16}\right)}{{\left(\dfrac{- 15}{16}\right)}^{2}}\right]$

$⇒ 225\left[{f}^{'}\left(\dfrac{1}{2}\right) - {f}^{"}\left(\dfrac{1}{2}\right)\right] = 225\left[\left(\dfrac{- 32}{15}\right) + \dfrac{4 \times 19 \times 16}{225}\right]$

$⇒ 225\left[{f}^{'}\left(\dfrac{1}{2}\right) - {f}^{"}\left(\dfrac{1}{2}\right)\right] = \left[- 480 + 1216\right] = 736$

$⇒ 225\left[{f}^{'}\left(\dfrac{1}{2}\right) - {f}^{"}\left(\dfrac{1}{2}\right)\right] = 736$

$⇒ 225\left[{y}^{'}\left(\dfrac{1}{2}\right) - {y}^{"}\left(\dfrac{1}{2}\right)\right] = 736$