What is the correct derivative for the quotient \frac{\sqrt{x}}{x^3+1}?

[Nano-Passion]

EDIT: I found the mistake, question is answered! Its funny because I spent 40+ minutes trying to get the right answer and looking for the mistake but typing it all out in latex helped me to find it! ​

Homework Statement

$$\frac{\sqrt{x}}{x^3+1}$$

The Attempt at a Solution

$$\frac{\sqrt{x}}{x^3+1}$$
$$\frac{d/dx (x^{1/2}(x^3+1))-d/dx(x^{3}+1)x^{1/2}}{(x^3+1)^2}$$
$$\frac{\frac{1/2x^{-1/2}(3x+1)3x^2(x^{1/2}}{(x^3+1)^2}}$$
$$\frac{\frac{x^3+1}{2\sqrt{x}}3x^{5/2}}{(x^3+1)^2}$$
$$\frac{x^3+3x^{5/2}+1}{2\sqrt{x}(x^3+1)^2}$$

But the correct answer is:

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

 

[Mark44]

Homework Statement

$$\frac{\sqrt{x}}{x^3+1}$$

The Attempt at a Solution

$$\frac{\sqrt{x}}{x^3+1}$$
$$\frac{d/dx (x^{1/2}(x^3+1))-d/dx(x^{3}+1)x^{1/2}}{(x^3+1)^2}$$

The line above is wrong.

You should have this:
$$\frac{d}{dx}\frac{\sqrt{x}}{x^3+1}$$
$$=\frac{(x^3+1)\cdot d/dx (x^{1/2})- x^{1/2}d/dx(x^{3}+1)}{(x^3+1)^2}$$
Can you continue?

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

But the correct answer is:

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