Question

What is the DERIVATIVE of:

ln x^2 * ln 6 -(product rule)

Answer 1

you are told to apply the product rule, so apply it

(fg)' = f'g + fg'

f = x^2
g = ln x

(I assume ln 6) is a typo.

Answer 2

Make the substitution u = x^2 and then use the product rule

d/dx (u) = d/dx(lnu)*du/dx
= 1/u*(du/dx)
= (1/x^2)(2x)
= 2/x

Don't forget the ln6 factor

Answer 3

oops - I see I missed the ln(x^2)

drwls caught it.
However, I still think ln6 is a typo, or there's no need for the product rule; just the chain rule.

So, if ln6 should be lnx then

f = ln x^2
g = ln x

note that f = 2 lnx
so, fg = 2ln^2(x)
but that is again moving away from the product rule. Good to check, though:

(2 ln^2 x)' = 2 * 2lnx * 1/x = 4/x ln x

Answer 4

the way it stands, there is no need for the product rule, since the ln6 is merely a constant

so
y = ln x^2 * ln 6
= 2lnx (ln6)
= 2ln6 (lnx)
dy/dx = 2ln6 (1/x)