Question

Derivative of Sqrt(x) - 3 / Sqrt (x) + 3
Quotient rule, or another way?

If you don't mind explaining how you got your answer, that would be awesome!

Derivative of Sqrt(x) - 3 / Sqrt (x) + 3
Quotient rule, or another way?

If you don't mind explaining how you got your answer, that would be awesome!

@Mathematics

Answer 1

\[\frac{d}{dx} \frac{\sqrt{x}-3}{\sqrt{x}+3}\]
is that what you are asking?
If so, Quotient rule would probably be your best bet.

Answer 2

When i do it, the top gets wiped out ...

Answer 3

your final answer is
\[\frac{3}{(\sqrt{x}+3)^2\sqrt{x}}\]

Answer 4

I will write it out , so If I made an error it can be pointed out.

\[((x^\left\{ .5 \right\}-3)(.5x^\left\{ -3/2 \right\})-(x^\left\{.5 \right\}+3)(.5x^\left\{ -3/2 \right\}))/(\sqrt{x}+3)^2 \]

Answer 5

\[(0.5x^{-1}-1.5x^{-3/2}-0.5x^{-1}+1.5x^{-3/2})/(\sqrt{x}+3)\]

Answer 6

^2*

Answer 7

the answer you provided wasn't accepted by the web work.

Answer 8

never mind it was the right answer, I forgot to plug in a value .... HUman error sorry. But can you show how you got it, if its not to much to ask ! :-)

Answer 9

yeah, I'm not sure what you are doing.

Quotient rule:
\[\frac{d}{dx}\frac{u}{v}=\frac{v\frac{du}{dx}-u\frac{du}{dx}}{v^2}\]
where:
u=x^(1/2)-3
and
v=x^(1/2)+3
which leads us to:
\[\frac{(\sqrt{x}-3)(\frac{d}{dx}(\sqrt{x}-3))-(\sqrt{x}+3)(\frac{d}{dx}(\sqrt{x}+3))}{(\sqrt{x}+3)^2}\]
as the first step.

Answer 10

\[\frac{(\sqrt x-3)(\frac{d}{dx}(-3)+\frac{d}{dx}(\sqrt x))-(\sqrt x +3)(\frac{d}{dx}(3)+\frac{d}{dx}(\sqrt x)) }{(\sqrt{x}+3)^2}\]
now that everything is separated we can start doing derivatives :D
\[\frac{(\sqrt x-3)(\frac{1}{2\sqrt x})+(\sqrt x+3)(\frac{1}{2 \sqrt x})}{(\sqrt x +3)^2}\]
simplifying a touch:
\[\large \frac{\frac{\sqrt x +3}{2 \sqrt x}-\frac{\sqrt x-3}{2\sqrt x }}{(\sqrt x +3 )^2}\]
from there you can evaluate and get to the answer I first said :D

Answer 11

thanks man