Question

find the derivatives d/dx (sqrt(x-1/x+1))

Answer 1

what a pain this is

Answer 2

you need the chain rule, the quotient rule, and a raft of algebra at the end to simplify
ready?

Answer 3

yupp

Answer 4

the derivative of \(\sqrt{f(x)}\) is
\[\frac{f'(x)}{2\sqrt{f(x)}}\]

Answer 5

here
\[f(x)=\frac{x+1}{x-1}\]so we need to find \(f'(x)\) using the quotient rule

Answer 6

This problem is easier done by logarithmic differentiation.

Answer 7

ack typo
\[f(x)=\frac{x-1}{x+1}\]so we need to find \(f'(x)\) using the quotient rule

Answer 8

the x-1 is on the top in my problem

Answer 9

ok :) just making sure

Answer 10

@eliassaab yes that is certainly true, but
a) i am willing to wager she has not gotten there yet and
b) if you leave the answer in that form it will probably not look like what they want

Answer 11

If you put
y=\left\frac {x-1}{x+1} \right)^{1/2}

Answer 12

\[
y=\left(\frac {x-1}{x+1} \right)^{1/2}
\]

Answer 13

\[\ln (y) =\frac 1 2 \ln (x-1) - \frac 1 2 \ln (x+1)
\]

Answer 14

in any case i have to run because the twilight zone is coming on, but you need
\[f'(x)\] which in this case is not too hard,
\[\frac{(x+1)-(x-1)}{(x+1)^2}=\frac{2}{(x+1)^2}\]

Answer 15

so your final job is to write
\[\frac{\frac{2}{(x+1)^2}}{2\sqrt{\frac{x-1}{x+1}}}\] and then have an algebra party

Answer 16

ok im sorry but im really confused now

Answer 17

\[
\frac {y'} y =\frac 1{2 (x-1)}-\frac 1{2 (x+1)} \\

y'= y \left ( \frac 1{2 (x-1)}-\frac 1{2 (x+1)} \right)
\]

Answer 18

i am sure @eliassaab way will work swifty land nicely, but the devil will still be in the algebra
if you know logarithmic diff, use it

Answer 19

i dont think i really do so can we start from the top and go through it pplleeassee

Answer 20

look at the first post i wrote to give the idea of what you have to compute
i did the necessary computation
what is left is the algebra to make
\[\frac{\frac{2}{(x+1)^2}}{2\sqrt{\frac{x-1}{x+1}}}\] look more reasonable
hold on i need to log off and log on again
back in 2 minutes

Answer 21

Final simplified answer
\[
y'=\frac{1}{2}
\sqrt{\frac{x-1}{x+1}}
\left(\frac{1}{x-1}-\frac{1}{x+
1}\right)=\frac{\sqrt{\frac{x-1
}{x+1}}}{x^2-1}
\]

Answer 22

An equivalent answer will be
\[y'=\frac{1}{\sqrt{\frac{x-1}{x+1}}
(x+1)^2}

\]

Answer 23

if you got this, let me know
if you want to start at the top we can do that too

Answer 24

i think i got it

Answer 25

\[d/dx \left[ \left( x ^{2+1} \right)\sin \sqrt{x} \right]\]

Answer 26

2+1?

Answer 27

supposed to be \[x^{2}+1\]

Answer 28

does that make sense?