Homework Help: Easy second derivative question

1. Nov 29, 2007

demersal

[SOLVED] easy second derivative question

1. The problem statement, all variables and given/known data
"If y^2 - 3x =7 what is the second derivative?"

2. Relevant equations

A) -6/7y^3
B) -3/y^3
C) 3
D) 3/2y
E) -9/4y^3

3. The attempt at a solution

I got the first derivative to be: 3/2y
Second derivative: -6/(4y^2)

I cannot for the life of me figure out what I did wrong, but I know it's something since that's not a choice. I've been messing up a lot lately, so I think I'm making some sort of vital error. A step by step explanation would be appreciated!

You guys are so awesome here, thank you for all that you do.

2. Nov 29, 2007

Dick

If y'=3/(2y)=(3/2)*y^(-1) (and it is) then y''=(3/2)*(-1)*y^(-2)*y'. Notice the y' on the right side coming from the chain rule. Now substitute your previous expression for y' for that y'.

3. Nov 29, 2007

rock.freak667

well the first derivative is $\frac{dy}{dx}=\frac{3}{2y}=\frac{3}{2}y^{-1}$
diff. w.r.t x again

$$\frac{d^2y}{dx^2}=\frac{-3}{2}y^{-2}\frac{dy}{dx}$$

which should give....$$\frac{d^2y}{dx^2}=\frac{-3}{2}y^{-2}\frac{3}{2}y^{-1}$$

I think you multiplied wrongly

4. Nov 29, 2007

demersal

Dick, you are beyond wonderful. I kept messing up because I wasn't subbing for y' properly, but now I see the trick to use the first derivative.

THANK YOU!