Question

Let f(x) = logbase6 (4x^2-2x+6) find f'(x)
So, I know that d/dx logbasea = 1/xlna
so I thought it would start by looking like this
1/((4x^2-2x+6)ln6) Then from here I thought you might have to use the Quotent rule which would look like this
(4x^2-2x+6)(0)-1(8x-2)/((4x^x-2x+6)ln6)^2
which the top simplified would be
-8x+2/((4x^2-2x+6)ln6)^2
but, now I am stuck please see if what I am doing is correct and help me figure out what is next. thanks.

Answer 1

To be clear,
\[\frac{d}{dx}\log_{6}(4x^{2}-2x+6)\]
Is what you're trying to find?

Answer 2

Yes

Answer 3

You did well up to
\[\frac{1}{(4x^{2}-2x+6)\ln6}\]
But you will multiply this to the derivative of
\[4x^{2}-2x+6\]
and not to the derivative of
\[\frac{1}{(4x^{2}-2x+6)\ln6}\]

So for this particular problem you don't have to worry about the quotient rule... :)

Answer 4

so you would just do the 1/(4x^2-2x+6)ln6 *8x-2?

Answer 5

Yes, that looks about right, and you're done!

Answer 6

Thanks!

Answer 7

No problem