5. c) If g(x) = 1 / square root (2x+1), use the definition of the derivative to find g prime (1).
I would greatly appreciate a short walk through on this number. I've divided the steps into 5.
Step 1 was to arrange everything so it looks like the formula
g(x+h) - g prime(x) / h
Step 2 was to put everything on common denominator.
Step 3 rationalize - this is where I get stuck
The teacher went over it but just "said the nominator becomes -2h". I tried cracking this and cannot get -2h. There must be something wrong with my algebra, I've tried many many times and by substituing x=1 I still can't.
Step 4 Is to simplify and we should end up with -1 / 3^(3/2).
I am able to get -2h and of course the h cancels on steps 5 to 6.
Now the transition from step 6 to 7 is odd. I notice we multiply outer with outer and inner with inner to obtain what you have on the step 7 denominator, but how come there is no sign of the inner*outer and outer*inner multiplications?
Okay I understand how we get -2 / (2x+1) (2 square root 2x+1) but I can't get the final answer you get.
I know it has something to do with the second term in the denominator but I can't simplify it into the first term to obtain an exponent of 3/2.
I've tried substituting for one (g prime (1)). and get:
Step 1) -2 / square root 3 * square root 3 ( square root 3 + square root 3).
Step 2) -2 / 3 ( square root 3 + square root 3)
This is where I'm stuck.
Yes thank you, I did not see that. I was able to solve it by altering (square root 3 + square root 3) into (3+3)^1/2 and then (3x2)^1/2.
My algebra is incredibly rusty but if what I did is correct then the 2's cancel and we get
-1 / 3 * 3^1/2 = -1 / 3^3/2
Whew!
I do prefer your method. Thanks for the help!