Differentiation - Quotient Rule Help!!?

y= u/v then dy/dx = vu'-uv'/v^2 (u and v functions of x)

so you need to find the derivative of a logarithm (ln).

ie what is f(x) in the expression; ln(ax+b) = integral f(x) dx?

Yeah I'm aware that dy/dx of ln(ax+b) = a/ax+b

What I get is

v= 4lnx-3 dv/dx= 4/x-3
u= 4lnx+3 du/dx= 4/x+3

((4/x+3)4lnx-3 - (4/x-3)4lnx+4) / (4lnx-3 )^2


I dont know where to go from there.

bearing in mind I havn't checked anything:

a/x + b/y = (ay+xb)/xy

where (4/x+3)ln(x-3) =a and x=1 and rather than a + sign you have a - sign, try then simplifying it using log' rules such as ln(ab)=ln(a)+ln(b) (or whatever those rules are

In this question though when would I need to use the log rules?


The mark scheme says this, but I dont understand what its means exactly.

In this question though when would I need to use the log rules?


The mark scheme says this, but I dont understand what its means exactly.

Methinks the question is supposed to be, in LaTex form:

$[br]\dfrac{4\ln (x) -3}{4\ln (x) +3}[br]$

In which case, you should be able to get the correct answer.

y = 4ln(x-3) / 4ln(x+3)
let u=4ln(x-3) u'=4/(x-3)
v=4ln(x+3) v'=4/(x+3)

dy/dx = (vu'-uv')/v^2

= (4/x-3)ln(x+3) - (4/x+3)ln(x-3)
-------------------------------------
(4ln(x+3))^2

= 4 ((1/x-3)ln(x+3) - (1/x+3)ln(x-3))
------------------------------------------
4^2 . ln(x+3)^2

= (1/x-3)ln(x+3) - (1/x+3)ln(x-3)
---------------------------------------
4ln(x+3)^2

try using the addition of fraction rule I told you in my last post

= (x+3)ln(x+3) -(x-3)ln(x-3)
-----------------------------------
4ln(x+3)^2

= (x+3)ln(x+3) - (x-3)ln(x-3)
--------------------------------
12ln(x+3)

= (x+3) - (x-3) ln(x-3 - (x+3))
------ -----
12 12

something tells me I havn't done this right, or at least the way you need it.

Honestly, I think you're right. I was thinking that it was f(x-3) when its actually f(x)-3.
Thank you!!!