Question

f(s) = (√s - 1) / (√s + 1)
f'(s) = ?

Answer 1

It'll be easier to change the square root to s^(1/2)... then use the quotient rule.

Answer 2

ugly algebra involved for sure
\[(\frac{f}{g})'=\frac{gf'-fg'}{g^2}\] use
\[f(x)=\sqrt{x-1},f'(x)=\frac{1}{2\sqrt{x-1}}\]
\[g(x)=\sqrt{x+1},g'(x)=\frac{1}{2\sqrt{x+1}}\] and make the replacements

Answer 3

btw it is good to remember that
\[\frac{d}{dx}\sqrt{x}=\frac{1}{2\sqrt{x}}\] it comes up frequently and it is always the same, sort of like memorizing 7 times 8

Answer 4

Sometimes it's easier to find the derivative by making the question look a little simpler for yourself. If it makes it easier and you're more comfortable with the expansion with the multiplying rule, instead of seeing \[\sqrt{s}\] think of it as \[s^{1/2}\]

Make it \[f(s) = (s^{1/2} -1) (s ^{1/2} +1)^{-1}\]

Answer 5

so what is s^1/2 times 1/2 s^-1/2 ??

Answer 6

i completely disagree

Answer 7

f(s)=(s1/2−1)(s1/2+1)−1 would mean i would have to use the chain rule, and i want to solve this using the quotient rule

Answer 8

Then you would follow what satellite used

Answer 9

also i am assuming you meant
\[\frac{\sqrt{x-1}}{\sqrt{x+1}}\] although i could be wrong

Answer 10

*Chain rule, the name slipped my mind for a second

Answer 11

no, only the sq. roots of x in each of those

Answer 12

oooh in that case it is much easier

Answer 13

\[f(x)=\sqrt{x}-1,f'(x)=\frac{1}{2\sqrt{x}}\]
\[g(x)=\sqrt{x}+1,g'(x)=\frac{1}{2\sqrt{x}}\]

Answer 14

\[\frac{(\sqrt{x}+1)\times \frac{1}{2\sqrt{x}}-(\sqrt{x}-1)\times \frac{1}{2\sqrt{x}}}{(\sqrt{x}+1)^2}\]

Answer 15

\[f(s) = \frac{s^{1/2}-1}{s^{1/2}+1}\]
\[f'(s) = \frac{1}{(s^{1/2}+1)^2s^{1/2}}\]

Answer 16

Alright so here is the answer using only the division rule:

u = s^1/2 - 1
u' = (1/2)s^-(1/2)
v = s^1/2 + 1
v' = (1/2)s^-(1/2)

f'(s) = (s^1/2 + 1)((1/2)s^-(1/2)) - (s^1/2 - 1)((1/2)s^-(1/2)) / (√s + 1)²

= (1/2) + (1/2s^-(1/2)) - ((1/2) - ((1/2s^-(1/2) / (√s + 1)²

= (1/2s^-(1/2) + (1/2s^-(1/2) / (√s + 1)²

= s^-(1/2) / (√s + 1)²

= 1/ √s(√s+1)²

Answer 17

multiply top and bottom by
\[\frac{1}{2\sqrt{x}}\] to clear the fractions and get
\[\frac{\sqrt{x}+1-(\sqrt{x}-1)}{2\sqrt{x}(\sqrt{x}+1)^2}\]
\[=\frac{2}{2\sqrt{x}(\sqrt{x}+1)^2}\]
\[=\frac{1}{\sqrt{x}(\sqrt{x}+1)^2}\]

Answer 18

you got it!

Answer 19

there is in fact a somewhat snappier way to do it using the chain rule, but it might not be worth the effort

Answer 20

yes it took me a long time to realize that when you multiply s^1/2 by 1/2s^-(1/2) that the answer is just 1/2 because you add the exponents together so (1/2) + -(1/2) = 0 and then s^0 is just 1

so s^1/2 * 1/2s^-(1/2) = 1/2

Answer 21

yes it is true. it might be easier to do this problem
\[\frac{d}{dx}\frac{x-1}{x+1}\] and get
\[\frac{2}{(x+1)^2}\] then replace x by
\[\sqrt{x}\] and multiply by
\[\frac{1}{2\sqrt{x}}\] but that requires some thinking which might not be clear

Answer 22

The hardest part of this question was knowing where to start and how to rearrange the original equation to work step by step in order to simplify things