If y(x)=xˣ,x > 0, then y''(2)-2y'(2) is equal to

Solution

Maths sir · 5 steps

Take logs to get y', then differentiate y' = y(1+ln x) once more with the product rule.

1. First derivative. Write ln y = x ln x, where ln means logₑ, and differentiate both sides: y'/y = ln x + 1. So y' = xˣ(1+ln x).
2. Second derivative. Use the product rule on y' = y(1+ln x). Then replace the y' that appears with y(1+ln x).

   y'' = y'(1+ln x) + y/x = xˣ(1+ln x)² + xˣ⁻¹

3. Put in numbers. At x = 2, y(2) = 4. So y'(2) = 4(1+ln 2) and y''(2) = 4(1+ln 2)² + 2.
4. Combine. y''(2) - 2y'(2) = 4 + 8 ln 2 + 4(ln 2)² + 2 - 8 - 8 ln 2. The 8 ln 2 terms cancel, and the constants give 4 + 2 - 8 = -2.
5. So. y''(2) - 2y'(2) = 4(logₑ 2)² - 2, which is option (d).

The idea to keep. For xˣ type functions, write y' = y · (something) and keep y as y while differentiating again. The product rule then hands you y'' neatly.

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