Can someone please help me out... I've been struggling with this question for a while. I'm decent at doing just the chain rule or just the quotient rule but for some reason I can't do them together. The question is:
Find the derivative of:
1/ square root of 2x-3
One part that really confuses me is when i do the quotient rule:
f' (x) g (x) - g' (x) f (x) = 0 - 1x-3^-1/2 (1) / g(x)^2 = ???
I can't seem to figure out g(x) squared... G (x) is 2x-3^-1/2, and if we do that squared would it be just 2x-3 ?? (1/2 x 2 = 1) ?
if that's the case I have something like:
-x-3^-1/2 / (2x-3) squared
= square root (-x-3) / (2x-3) squared...
as you can see it's a jumbled mess.. If someone can provide me an answer step by step that'll really help. I tried it but can't seem to get it. Thanks in advance!