# Thread: Differentiating the root of a quotient

1. ## Differentiating the root of a quotient

How do you differentiate $\sqrt{\left ( \frac{1-x}{1+x}\right )}$ with respect to x?

I've tried using $\frac{dy}{du}\times \frac{du}{dx}$ with $u=\left ( \frac{1-x}{1+x}\right )$ and $u=(1-x)(1+x)^{-1}$ but neither seems to get the right answer.

2. Originally Posted by djnorris2000
How do you differentiate $\sqrt{\left ( \frac{1-x}{1+x}\right )}$ with respect to x?

I've tried using $\frac{dy}{du}\times \frac{du}{dx}$ with $u=\left ( \frac{1-x}{1+x}\right )$ and $u=(1-x)(1+x)^{-1}$ but neither seems to get the right answer.

You can consider that is...

... so that the standard solution for the derivative of the ratio of two functions can be applied...

Kind regards

$\chi$ $\sigma$

3. Thanks for that. I wonder if you'd mind checking my working as I still don't seem to be arriving at the right answer?

$y=\sqrt{\left ( \frac{1-x}{1+x} \right )}$

Let $u=\sqrt{1-x}$ and $v=\sqrt{1+x}$

Therefore

$\frac{du}{dx}=\frac{1}{2}(1-x)^{-\frac{1}{2}}\times -1=\frac{-1}{2(1-x)^{\frac{1}{2}}}$

and
$\frac{dv}{dx}=\frac{1}{2}(1+x)^{-\frac{1}{2}}\times 1=\frac{1}{2(1+x)^{\frac{1}{2}}}$

The differential of a quotient is

$\frac{v\frac{du}{dx}\times u\frac{dv}{dx}}{v^{2}}$

So
$\frac{(1+x)^\frac{1}{2}\frac{-1}{2(1-x)^\frac{1}{2}}\times (1-x)^\frac{1}{2}\frac{1}{2(1+x)^\frac{1}{2}}}{((1+x) ^\frac{1}{2})^{2}}$

Becomes

$\frac{\frac{-(1+x)^{\frac{1}{2}}}{2(1-x)^{\frac{1}{2}}}\times \frac{(1-x)^\frac{1}{2}}{2(1+x)^{\frac{1}{2}}}}{(1+x)}$

Becomes

$\frac{\frac{-2(1+x)\times 2(1-x)}{4(1+x)^{\frac{1}{2}}(1-x)^{\frac{1}{2}}}}{1+x}$

Becomes

$\frac{-4(1+x)(1-x)}{4(1+x)^{\frac{3}{2}}(1-x)^{\frac{1}{2}}}$

Becomes

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

Becomes

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

4. Or, the way I would be inclined to do it:
$\sqrt{\frac{1- x}{1+ x}}= \left(\frac{1- x}{1+ x}\right)^{1/2}$

The derivative of that is
$\frac{1}{2}\left(\frac{1- x}{1+ x}\right)^{-1/2}= \frac{1}{2}\left(\frac{1+ x}{1- x}\right)^{1/2}$
times the derivative of the fraction
$\frac{1- x}{1+ x}$
which can be done using the quotient rule.

Oh, ho! Using "[ tex ]" and "[ /tex ]" (without the spaces) now gives the LaTeX!

5. This was the way I first tried solving it but I wasn't arriving at the right answer. I tried this way on Chisigma's advice but I'm still not arriving at the right answer.

6. the formula for a quotient (u/v)' is (vu' - uv')/v^2 NOT (vu')(uv')/v^2. i don't see the right formula in your post.

7. Originally Posted by Deveno
the formula for a quotient (u/v)' is (vu' - uv')/v^2 NOT (vu')(uv')/v^2. i don't see the right formula in your post.
"Lo dee-high minus high dee-low over the square of what's below" is my mnemonic. Incidentally, while you never technically need the quotient rule, I find it helpful because it eliminates the need to get a common denominator and add fractions later. More calculus now = less algebra later.

8. That is one epic mnemonic.

9. Originally Posted by djnorris2000
How do you differentiate $\sqrt{\left ( \frac{1-x}{1+x}\right )}$ with respect to x?
\displaystyle \begin{aligned} f(x) = \sqrt{\frac{1-x}{1+x}} & \Rightarrow \ln(f(x)) = \ln\sqrt{\frac{1-x}{1+x}} = \frac{1}{2}\ln\left(1-x\right) -\frac{1}{2}\ln\left(1+x\right) \Rightarrow \frac{f'(x)}{f(x)} = \cdots \end{aligned}

10. I always think deriviative of the thing in the bracket (which in this case is found by the quotient rule ) multiplied by what you would normally do in differentiation. Ok not worded very well. I end up with -1/((1+x^2)*sqrt((1-x)/(1+x))). Hmm maybe too many brackets spoil the broth

11. the thing "inside the root" is: (1-x)/(1+x), which has the derivative [(1+x)(-1) - (1-x)(1)]/(1+x)^2 = -2/(1+x)^2.

as HallsofIvy pointed out, (√u)' = 1/(2√u) = (1/2)√(1/u), which in this case becomes (1/2)√((1+x)/(1-x)).

combining the two the 1/2 and 2 cancel, and we may write (1+x)^2 as √(1+x)^4, making the answer:

-1/(√[(1-x)(1+x)^3]) ( will attempt to put this in tex:

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

hopefully it will display right), which integrates correctly according to mathematica.

12. Originally Posted by Ackbeet
"Lo dee-high minus high dee-low over the square of what's below" is my mnemonic. Incidentally, while you never technically need the quotient rule, I find it helpful because it eliminates the need to get a common denominator and add fractions later. More calculus now = less algebra later.
my high-school teacher taught it as:

"bottom dee top less top dee bottom over bottom squared."

13. Thanks for the help, I'm kicking myself over the quotient formula as that's what the previous chapter of the book was on, I must have misread it when flicking backwards.

Anyway, I think I have an answer now...

$y=\sqrt{\left ( \frac{1-x}{1+x} \right )}$

Let $u=\frac{1-x}{1+x}$

Therefore
$\frac{dy}{du}=\frac{1}{2}u^{-\frac{1}{2}}=\frac{1}{2}\left ( \frac{1-x}{1+x} \right )^{-\frac{1}{2}}=\frac{1}{2}\times \frac{1}{\left ( \frac{1-x}{1+x} \right )^{\frac{1}{2}}}=\frac{1}{2}\times \frac{1}{\frac{\left ( 1-x \right )^{\frac{1}{2}}}{\left ( 1+x \right )^{\frac{1}{2}}}}=\frac{1}{2}\times \frac{1}{\left ( 1-x \right )^{\frac{1}{2}}\left ( 1+x \right )^{\frac{1}{2}}}=\frac{1}{2\left ( 1-x \right )^{\frac{1}{2}}\left ( 1+x \right )^{\frac{1}{2}}}$

And
$\frac{du}{dx}=\frac{-(1+x)-(1-x)}{(1+x)^{2}}=\frac{-2-2x}{(1+x)^{2}}=\frac{-2(1+x)}{(1+x)^{2}}=\frac{-2}{1+x}$

Therefore
$\frac{dy}{du}\times \frac{du}{dx}=\frac{1}{2\left ( 1-x \right )^{\frac{1}{2}}\left ( 1+x \right )^{\frac{1}{2}}}\times \frac{-2}{1+x}=\frac{-2}{2\left ( 1-x \right )^{\frac{1}{2}}\left ( 1+x \right )^{\frac{3}{2}}}=\frac{-1}{\left ( 1-x \right )^{\frac{1}{2}}\left ( 1+x \right )^{\frac{3}{2}}}$

That's giving me the same answer as the book but I'm not sure if I've worked out $\frac{dy}{du}$ correctly as while $\frac{1}{2}\times \frac{1}{\frac{\left ( 1-x \right )^{\frac{1}{2}}}{\left ( 1+x \right )^{\frac{1}{2}}}}=\frac{1}{2}\times \frac{1}{\left ( 1-x \right )^{\frac{1}{2}}\left ( 1+x \right )^{\frac{1}{2}}}=$ looks right it seems very counter intuitive to me.

Bah, Deveno just typed the whole thing out while I was struggling with LaTeX and answered the question. Thanks very much.

14. if you worked through it, did the work, and arrived at an answer that agrees with someone else's, which then makes sense to you...then that is a good thing

15. Originally Posted by Ackbeet
"Lo dee-high minus high dee-low over the square of what's below" is my mnemonic. Incidentally, while you never technically need the quotient rule, I find it helpful because it eliminates the need to get a common denominator and add fractions later. More calculus now = less algebra later.
I could sing this in the shower!