If y = y(x) is an implicit function of x such that logₑ(x + y) = 4xy, then d²y/dx² at x = 0 is equal to ___________.

Solution

Maths sir · 4 steps

Rewrite it as x + y = e⁴ˣʸ, find y and y' at x = 0 first, then differentiate once more.

1. Value of y. Put x = 0: ln y = 0, so y = 1. Then remove the log by writing the relation as x + y = e⁴ˣʸ.
2. First derivative. Differentiate both sides: 1 + y' = e⁴ˣʸ(4y + 4xy'). Since e⁴ˣʸ = x + y, this becomes 1 + y' = (x + y)(4y + 4xy'). At x = 0, y = 1: 1 + y' = 1 × 4 = 4, so y' = 3.

   y'' = (1 + y')(4y + 4xy') + (x + y)(4y' + 4y' + 4xy'')

3. Second derivative. This is the product rule applied to (x + y)(4y + 4xy'). Note that d/dx(4xy') = 4y' + 4xy''. At x = 0: y'' = (4)(4) + (1)(8 × 3) = 16 + 24.
4. So. d²y/dx² at x = 0 is 40.

The common slip: 28. Treating x in 4xy' as a constant, so its derivative is only 4xy'', loses a 4y' term and gives 16 + 12 = 28, not 40. The product rule gives 4y' + 4xy''.

The idea to keep. For a derivative at one point, find y first, then y', then y'', putting in known values at each stage. Swapping e⁴ˣʸ back to x + y keeps the algebra short.

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