# 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!