C3 quotient rule help!

Where have i gone wrong here ? i tried to check it but my checking suggests whst i have done is wrong...

Hi! It's that step in the middle where you suddenly get rid of (x+3) from the quotient. It should be:

$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{ \frac{1}{2}(x+3)^{1/2}(x+1)^{-1/2}-\frac{1}{2}(x+3)^{-1/2}(x+1)^{1/2} }{(x+3)}$$= \frac{1}{2}(x+3)^{-1/2}(x+1)^{-1/2}-\frac{1}{2}(x+3)^{-3/2}(x+1)^{1/2}$

You can do the rest!

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Also that should be -1/2 on the index there, not -3/2 ;D

Tested the result on a function plotter; seems okay as it is!

Hi! It's that step in the middle where you suddenly get rid of (x+3) from the quotient. It should be:

$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{ \frac{1}{2}(x+3)^{1/2}(x+1)^{-1/2}-\frac{1}{2}(x+3)^{-1/2}(x+1)^{1/2} }{(x+3)}$$= \frac{1}{2}(x+3)^{-1/2}(x+1)^{-1/2}-\frac{1}{2}(x+3)^{-3/2}(x+1)^{1/2}$

You can do the rest!

Hi, thanks for helping. I have tried this again butstill didnt get it right. I dont get why it is -1/2 not +1/2 since there are two negative signs thanks!

You're doing it wrong.

In the quotient rule, let's say for example in $\frac{x+2}{x}$, this is of the form $\frac{v}{u}$.

You're going to set $v = x+2 \Rightarrow \frac{\mathrm{d}v}{\mathrm{d}x} = 1$ and $u = x \Rightarrow \frac{\mathrm{d}u}{\mathrm{d}x}$.

Note that we are not setting $u = x^{-1}$.

The whole point of the quotient rule is that you avoid that, so in your answer you should have $u = (x+3)^{1/2}$.

This gets you $y = \frac{v}{u} = \frac{\sqrt{x+1}}{\sqrt{x+3}}$ which is good, that's what you want to differentiate.

If we insist on using your definition, then we get $y = \frac{(x+1)^{1/2}}{(x+1)^{-1/2}} = \frac{\sqrt{x+1}}{\frac{1}{\sqrt{x+3}}} = \sqrt{x+1}\sqrt{x+3}$, nothing at all like your original function, you get the drift?

Thanks, i figured something was wrong with it. I tried using just (x+3) 1/2 but it still doesnt give me the answer. My book says its y=u/v so i dont know if thats why...

I agree up to the second = after dy/dx, after that, your simplification is wrong.

Pay attention to your indices. It should be:

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

Which gets you:

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

Now when you factorise, be very careful, pay attention to the indices.

Why did you multiply du/dx and dv/dx? I thought it is v* du/dx - u *dv/dx...

Hi, thanks for helping. I have tried this again butstill didnt get it right. I dont get why it is -1/2 not +1/2 since there are two negative signs thanks!

Careful, don't get confused with the notation! It seems as though you could do with an explanation as to how the quotient rule works. lets say $f(x)$ is what we want to differentiate. It's more conventional to express the quotient as $\frac{u}{v}$ (which is what I'll do here) rather than $\frac{v}{u}$ but it really doesn't matter as long as you stay consistent. The quotient rule says that if $f(x)=\frac{u}{v}$, then the derivative of $f(x)$, $f'(x)$ is given by
$f'(x)=\frac{vu'-uv'}{v^2}$
If you're keen to derive this and see why then you can do it from the product rule using $\frac{u}{v}=uv^{-1}$.

If we let $f(x)=\frac{x+1}{x+3}$, we see it's already in the form of a quotient (pretty much a fraction), with $u=(x+1)^{1/2}$ and $v=(x+3)^{1/2}$. The derivatives of each of these:
$u'=\frac{1}{2} (x+1)^{-1/2}$
$v'=\frac{1}{2} (x+3)^{-1/2}$
Plugging these into the formula for the derivative, we have

$= \frac{1}{2}(x+3)^{-1/2}(x+1)^{-1/2}-\frac{1}{2}(x+3)^{-3/2}(x+1)^{1/2}$
and after this point, the only thing left to do is to manipulate the answer into the form you need. It all comes with practice!