Differentiation Question 5
Question 5 - 2024 (29 Jan Shift 2)
Let $y=\log _e\left(\frac{1-x^{2}}{1+x^{2}}\right),-1<x<1$. Then at $x=\frac{1}{2}$, the value of $225\left(y^{\prime}-y^{\prime \prime}\right)$ is equal to
(1) 732
(2) 746
(3) 742
(4) 736
Show Answer
Answer (4)
Solution
$y=\log _e\left(\frac{1-x^{2}}{1+x^{2}}\right)$
$\frac{d y}{d x}=y^{\prime}=\frac{-4 x}{1-x^{4}}$
Again,
$\frac{d^{2} y}{d x^{2}}=y^{\prime \prime}=\frac{-4\left(1+3 x^{4}\right)}{\left(1-x^{4}\right)^{2}}$
Again
$y^{\prime}-y^{\prime \prime}=\frac{-4 x}{1-x^{4}}+\frac{4\left(1+3 x^{4}\right)}{\left(1-x^{4}\right)^{2}}$
at $x=\frac{1}{2}$,
$y^{\prime}-y^{\prime \prime}=\frac{736}{225}$
Thus $225\left(y^{\prime}-y^{\prime \prime}\right)=225 \times \frac{736}{225}=736$
