Calculus: dy/dx= ?
(d^2y)/(dx^2)= ?
d^2y/dx^2= 0 at point (x, y)=?
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ln(5y) = 4xy

Differentiate both sides with respect to x, taking y = y(x) :

d/dx [ln(5y)] = d/dx [4xy]

1/(5y) d/dx [5y] = 4y + 4x dy/dx

5/(5y) dy/dx = 4y + 4x dy/dx

1/y dy/dx = 4y + 4x dy/dx

(1/y - 4x) dy/dx = 4y

dy/dx = 4y / (1/y - 4x)

Since y = 0 is outside the domain (ln(0) is undefined), we can multiply the right side by y/y :

dy/dx = 4y ² / (1 - 4xy)

To compute the second derivative, differentiate both sides again:

d/dx [dy/dx] = d/dx [4y ² / (1 - 4xy)]

d²y/dx ² = ((1 - 4xy) d/dx [4y ²] - 4y ² d/dx [1 - 4xy]) / (1 - 4xy)²

d²y/dx ² = ((1 - 4xy) 8y dy/dx - 4y ² (0 - 4y - 4x dy/dx)) / (1 - 4xy)²

d²y/dx ² = ((8y - 32xy ²) dy/dx + 16y ³ + 16xy ² dy/dx) / (1 - 4xy)²

d²y/dx ² = ((8y - 16xy ²) dy/dx + 16y ³) / (1 - 4xy)²

Substitute the first derivative founder earlier:

d²y/dx ² = ((8y - 16xy ²) (4y ² / (1 - 4xy)) + 16y ³) / (1 - 4xy)²

d²y/dx ² = ((8y - 16xy ²) 4y ² + 16y ³ (1 - 4xy)) / (1 - 4xy)³

d²y/dx ² = (16y ³ (3 - 8xy)) / (1 - 4xy)³

The second derivative vanishes wherever the numerator vanishes (so long as 4xy ≠ 1) :

16y ³ (3 - 8xy) = 0

16y ³ = 0   or   3 - 8xy = 0

y = 0   or   8xy = 3

As pointed out earlier, y = 0 is outside this function's domain, so we're left with points on the curve

y = 3/(8x)

Substituting this into the original function gives

ln(5 (3/(8x))) = 4x (3/(8x))

ln(15 / (8x)) = 12/8

ln(15) - ln(8x) = 3/2

ln(8x) = ln(15) - 3/2

8x = exp(ln(15) - 3/2)

8x = 15 exp(-3/2)

x = 15/8 exp(-3/2)   →   y = 1/5 exp(3/2)

or approximately (0.4184, 0.8963).

(where exp(x) = e ˣ)

linear pair postulates