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)