If loge (x+y) = 4xy. Find (d2y)/(dx2) at x = 0.

Solution:

Answer: 40

loge (x+y) = 4xy

or 1/((x + y)) [1 + (dy/dx)] = 4[x (dy/dx) + y]

or 1 + (dy/dx) = 4(x + y) [x (dy/dx) + y]⋯(i)

If x = 0, then y = 1

From (i)

1 + dy/dx = 4(0 + 1)[0 + 1] = 4

Or dy/dx = 3

From (i), again differentiate w.r.t. x

0 + (d2 y)/(dx2 ) = 4(x + y)[x (d2y)/(dx2 ) + 2 (dy/dx)] + 4[x (dy/dx) + y](1 + (dy/dx))

At x = 0, y = 1, dy/dx = 3

(d2 y)/(dx2) = 4(0 + 1)[0 + 2×3]+4[0 + 1](1 + 3) = 40

Therefore, (d2 y)/(dx2 ) = 40

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