Applications of Derivatives
Higher Order Derivatives — Logarithmic Function
nta_pyq_2024_jan
Grade 12
Question:
Let $y=\log_e\left(\dfrac{1-x^2}{1+x^2}\right)$, $-1<x<1$. Then at $x=\dfrac{1}{2}$, the value of $225(y'-y'')$ is equal to
Step-by-Step Solution
Key Concept: Differentiate $y=\ln(1-x^2)-\ln(1+x^2)$ to get $y'$. Differentiate again for $y''$. Then compute $y'-y''$ at $x=1/2$ and multiply by 225.
$y'=\frac{-4x}{1-x^4}$, $y''=\frac{-4(1+3x^4)}{(1-x^4)^2}$. At $x=1/2$: $1-x^4=15/16$. $y'=\frac{-2}{15/16}=\frac{-32}{15}$... computing $225(y'-y'')=736$.
Correct Answer: 4