Let y=log ₑ((1-x²)/(1+x²)),-1 < x<1 . Then at x=1/2 , the value of 225(y′-y^(′ ′)) is equal to

Solution

Maths sir · 4 steps

Split the log into two logs first, then the derivatives come out clean and short.

1. Split the log. Write y = ln(1-x²) - ln(1+x²). Each piece needs only one use of the chain rule.
2. First derivative. y' = -2x/(1-x²) - 2x/(1+x²) = -2x[(1+x²)+(1-x²)]/(1-x⁴) = -4x/(1-x⁴). At x = 1/2, 1-x⁴ = 15/16, so y' = -2 × 16/15 = -32/15.

   y'' = d/dx(-4x/(1-x⁴)) = -4 ((1-x⁴) + 4x⁴)/(1-x⁴)² = -4(1+3x⁴)/(1-x⁴)²

3. Second derivative value. At x = 1/2, 1+3x⁴ = 19/16 and (1-x⁴)² = 225/256. So y'' = -4 × 19/16 × 256/225 = -1216/225.
4. So. y' - y'' = -480/225 + 1216/225 = 736/225, and 225(y' - y'') = 736, option (b).

The idea to keep. Whenever you see the log of a quotient, break it into a difference of logs before differentiating. Combining the fractions early, as with 1-x⁴ here, keeps the second derivative manageable.

Answer from the NTA final answer key. Solution by PrepWiser.