# Thread: Help with using the quotient rule on a derivative?

1. ## Help with using the quotient rule on a derivative?

I don't know why, but for some reason I absolutely cannot get the answer to this derivative:

h(x) = sqrt(x)/x^3+1

I got as far as writing everything out and trying to simplify it:
(1/2x^-1/2)(x^3+1)-(x^1/2)(3x^2)/(x^3+1)^2
Simplifying:
(1/2)x^-3/2+(1/2)x^-(1/2)-3x/x^3+1

Somehow, I'm supposed to make this equal
1-5x^3/2sqrt(x)(x^3+1)^2

I think I might be making a simple algebra mistake, but I can't find it, so if someone could show me the process to get to the answer, that would help immensely

2. Originally Posted by Ohoneo
I don't know why, but for some reason I absolutely cannot get the answer to this derivative:

h(x) = sqrt(x)/x^3+1

I got as far as writing everything out and trying to simplify it:
(1/2x^-1/2)(x^3+1)-(x^1/2)(3x^2)/(x^3+1)^2
Simplifying:
(1/2)x^-3/2+(1/2)x^-(1/2)-3x/x^3+1

Somehow, I'm supposed to make this equal
1-5x^3/2sqrt(x)(x^3+1)^2

I think I might be making a simple algebra mistake, but I can't find it, so if someone could show me the process to get to the answer, that would help immensely

$h'(x) = \frac{(x^3+1)\frac{1}{2\sqrt{x}} - \sqrt{x}(3x^2)}{(x^3+1)^2}$

multiply by $\frac{2\sqrt{x}}{2\sqrt{x}}$ ...

$h'(x) = \frac{(x^3+1) - 2x(3x^2)}{2\sqrt{x}(x^3+1)^2}$

$h'(x) = \frac{1 - 5x^3}{2\sqrt{x}(x^3+1)^2}$

3. Ah, thank you. I was getting too ahead of myself and forgetting to rationalize. Thanks