Using the Log Law `log\ a^n= n\ log\ a`, we can write:

y = ln(2x3x)2 = 2 ln(2x3x)


`u = 2x^3 − x`


`u’ = 6x − 1`

This gives us:


`x ≠ ±sqrt(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

` Âħsqrt(0.5), 0`

Its graph is as follows:

Graph log(2x^30x)^2

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 `-sqrt 0.5 < x < 0` and `x > sqrt0.5.`

Graph log(2x^30x)^2 - correct domain

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.