Question

Find the derivative of: f(x)=[ x^(1/2)+1]/[x^2 + 1]

Answer 1

\[\Large\bf\sf f(x)\quad=\quad \frac{x^{1/2}+1}{x^2+1}\]
We'll need to apply the Quotient Rule:\[\Large\bf\sf f'(x)\quad=\quad \frac{\color{royalblue}{(x^{1/2}+1)'}(x^2+1)-(x^{1/2}+1)\color{royalblue}{(x^2+1)'}}{(x^2+1)^2}\]

Answer 2

Does the setup make sense? We need the derivatives of the blue parts.

Answer 3

Yes, the setup makes sense. So it would become [1/2x^{-1/2}(x^2+1) - x^(1/2)+1)(2x)] / (x^2+1)^2 right?

Answer 4

Mmm ya looks good!

Answer 5

Ok, thank you. But I'm on confused on what to do after that lol.

Answer 6

The answer in the book is: (0.5x^{-1/2})[x^2+1-4x^{3/2}(x^{1/2})]/(x^2+1)^2 OR -3x^2-4x^{3/2} + 1 / 2x^{1/2}(x^2+1)^2

Answer 7

So the first answer, it looks like all they did was:
factor a (1/2)x^(-1/2) out of each term.

Answer 8

\[\Large\bf\sf f'(x)\quad=\quad \frac{\color{orangered}{(\frac{1}{2}x^{-1/2})}(x^2+1)-(x^{1/2}+1)\color{orangered}{(2x)}}{(x^2+1)^2}\]

Answer 9

So if we factor that first orange part out of each term.. we should get uhhhh..
So it will simply disappear from the first term, yes?
Then we need to alter the 2x to compensate.

Answer 10

It doesn't really matter though, dont' trust your book answers too much...
They tend to oversimplify things D:

Answer 11

\[\Large\bf\sf f'(x)\quad=\quad \color{orangered}{\left(\frac{1}{2}x^{-1/2}\right)}\frac{(x^2+1)-(x^{1/2}+1)\left(\frac{2x}{\left(\dfrac{1}{2}x^{-1/2}\right)}\right)}{(x^2+1)^2}\]

Answer 12

Okay then aha. Thank you!

Answer 13

Mmm I made that look a bit ugly :P
So dividing 2x and (1/2)x^(-1/2) gives us:\[\Large\bf\sf f'(x)\quad=\quad \color{orangered}{\left(\frac{1}{2}x^{-1/2}\right)}\frac{(x^2+1)-(x^{1/2}+1)\left(4x^{3/2}\right)}{(x^2+1)^2}\]

Answer 14

And then that's a little bit closer to what they have right? :\
blah lol

Answer 15

I think lol. The simplified ones the book has gets confusing.

Answer 16

ya :c