Expert: Paul Klarreich - 1/29/2007

Question
can you tell me how to solve the rest of this problem?:

find y'   
y=x(3x-4)^1/2



so far i've gotten to y'= 3/2x(3x-4)^-1/2 + (3x-4)^1/2  but i dont know where to go from there! THANKS!

Answer
Questioner:   Mara
Category:  Calculus
 
Subject:  calculus related rate
Question:  can you tell me how to solve the rest of this problem?:

find y'   
y=x(3x-4)^1/2

so far i've gotten to y'= 3/2x(3x-4)^-1/2 + (3x-4)^1/2  but i dont know where to go from there! THANKS!
.........................................  
Hi, Mara,

Where you go from there is to some simplifying.  Write your answer in radical form.  

You have, so far:
       3x
y = -------------  + (3x - 4)^1/2  
   2(3x - 4)^1/2

        3x
y = ---------------  + sqrt(3x - 4)
   2 sqrt(3x - 4)

Now use  2 sqrt(..) as the LCD to combine fractions.  The radical will disappear in the second numerator.

    3x + 2(3x - 4)
y = ----------------
    2 sqrt(3x - 4)

    3x + 6x - 8
y = ----------------
    2 sqrt(3x - 4)

       9x - 8
y = ----------------
    2 sqrt(3x - 4)

And that's about it.  You could rationalize the denominator, but it doesn't get you much.