If y = xx, prove that y(d2y/dx2) = y(dy/dx)2 - yx = 0.

y = xx
logy = xlogx
1/y (dy/dx) = 1 + logx
dy/dx = y(1 + logx) = y'— (1)
d2y/dx2 = y × 1/x + (1 + logx)dy/dx
d2y/dx2 = y/x + y'y.
y(d2y/dx2) = y(y/x + y'y) = y2/x + yy'2
From (1), y' = y(1+logx)
y(d2y/dx2) = y2/x + y[y(1+logx)]2 = y2/x + y3(1+logx)2
This doesn't directly lead to the given equation y(d2y/dx2) = y(dy/dx)2 - yx = 0. There might be an error in the original problem statement or the desired result. Let's re-examine the derivation:
y = xx
ln y = x ln x
differentiating both sides with respect to x:
(1/y) (dy/dx) = ln x + 1
dy/dx = y(ln x + 1)
differentiating again with respect to x:
d2y/dx2 = (dy/dx)(ln x + 1) + (y/x)
Substitute dy/dx = y(ln x + 1):
d2y/dx2 = y(ln x + 1)2 + y/x
Multiplying by y:
y(d2y/dx2) = y2(ln x + 1)2 + y2/x
This still doesn't match the given equation. The original problem statement or the target equation is likely incorrect.