If y(x) = ( xˣ )ˣ, x > 0, then d²x/dy² + 20 at x = 1 is equal to ____________.

Solution

Maths sir · 4 steps

First tidy (xˣ)ˣ into x^(x²). Then flip the derivatives with d²x/dy² = -y''/(y')³.

1. Simplify and log. (xˣ)ˣ = x^(x²), so ln y = x² ln x. Differentiating gives y'/y = 2x ln x + x, so y' = y (2x ln x + x).
2. Second derivative. Use the product rule: y'' = y' (2x ln x + x) + y (2 ln x + 3). At x = 1 we have ln 1 = 0 and y = 1, so y' = 1 and y'' = 1 × 1 + 1 × 3 = 4.
3. Invert the derivative. dx/dy = 1/y'. Differentiate this with respect to y by going through x: d/dx(1/y') × dx/dy = -y''/(y')² × 1/y'.

   d²x/dy² = -y''/(y')³ = -4/1³ = -4

4. So. d²x/dy² + 20 = -4 + 20 = 16.

The common slip: 20.25. Taking d²x/dy² as 1/y'' = 1/4 gives 20.25, not 16. The second derivative does not flip like the first one.

The idea to keep. d²x/dy² is not the reciprocal of d²y/dx². Always use d²x/dy² = -y''/(y')³, with derivatives taken in x.

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