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!​

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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}