y = ln(2x³ −x)² = 2 ln(2x³ −x)

Put

u = 2x³ − x

so

u' = 6x − 1

This gives us:

x ≠ ±√0.5,

x ≠ 0

NOTE: We need to be careful with the domain of this solution, as it is only correct for certain values of x.

The graph of y = ln(2x³ - x)² is defined for all x except

x = -√0.5, x = 0 and x = √0.5.

Its graph is as follows:

The graph of y = 2 ln(2x³ - x), however, is only defined for a more limited domain (since we cannot have the logarithm of a negative number.)

So we can only have x in the range -√0.5 < x < 0 and x > √0.5.

So when we find the differentiation of a logarithm using the shortcut given above, we need to be careful that the domain of the function and the domain of the derivative are stated.