# Find dy of x/( sqrt (3x + 6) )

1. Oct 23, 2012

### HellRyu

1. The problem statement, all variables and given/known data
I am doing my review for an exam Friday and the review shows only this: "find dy"

x/( sqrt(3x + 6) )

The review gives me the answer which is:
[(3x + 12)/2(3x + 6)^(3/2)]dx

2. The attempt at a solution

I started out by re-writing the sqrt as " (3x + 6)^1/2 " which I then proceeded to multiply the equation by the derivative of " (3x + 6) " and got:

3x/(3x + 6)^(1/2).

I'm stuck on how to proceed.

2. Oct 23, 2012

### Mentallic

Do you know the quotient rule for derivatives? Or equivalently, convert from

$$\frac{x}{(3x+6)^{1/2}}$$

to

$$x(3x+6)^{-1/2}$$

and use the product rule, setting $u=x$ and $v=(3x+6)^{-1/2}$

3. Oct 23, 2012

### phiiota

when you have functions like this, sometimes it's easier to write out your systems in general terms.

rewrite your equation as f(x)/g(x), or f(x)g(x)^-1, if you prefer, where f(x)=x and g(x)=√(3x+6). so now find the derivative of this. do it piece by piece until you have your solution.

4. Oct 23, 2012

### SammyS

Staff Emeritus
Hello HellRyu. Welcome to PF !

The question should have read something like:
Find dy, for $\displaystyle y=\frac{x}{\sqrt{3x + 6}}\ .$​
Of course, $\displaystyle dy=\frac{dy}{dx} dx=y'\,dx$

Basically, find y' and multiply by dx.

Use the quotient rule with $\displaystyle y=\frac{x}{(3x + 6)^{1/2}}\,,\$ or use the product rule with $\displaystyle y=x(3x + 6)^{-1/2}\ .$

5. Oct 24, 2012

### HellRyu

Hi, and thank you! :)

I started using the quotient rule which got me stuck at:

$$(-3x^{2} + 3x + 6)/ ( (3x + 6)^{3/2} )$$

6. Oct 24, 2012

### SammyS

Staff Emeritus
That's incorrect.

Show some intermediate steps.

7. Oct 24, 2012

### HellRyu

O.K. I thought it was. I have:

$$x/( (3x + 6)^{1/2} ) "f(x) / g(x)"$$

I then get $$"g(x)( f' (x) ) -( f(x)( g'(x) ) ) "$$

over

$$"g(x)^{2} "$$

I plug my f(x)'s and g(x)'s with their derivatives

$$(3x + 6)^{1/2} (1) - (x(3x) )$$

over

$$(3x + 6)^{3/2}$$
Is this correct so far? Or did I make a mistake somewhere?

8. Oct 24, 2012

### MarneMath

No, you need to use the chain rule on the 2nd part.

9. Oct 24, 2012

### HellRyu

Ha! You guys are genius', thank you all so much! I figured it out.

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