Using the Log Law `log\ a^n= n\ log\ a`, we can write:
y = ln(2x3 − x)2 = 2 ln(2x3 − x)
`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
` ±sqrt(0.5), 0`
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.