Question

can someone please help me

Answer 1

i would help if i could read it. are you looking for a derivative?

Answer 2

yes

Answer 3

is it
\[f(x)=\frac{x^2-1}{\sqrt{x^2-1}}\]?

Answer 4

oh no it looks like
\[f(x)=\frac{x^2-2}{\sqrt{x^2-1}}\]

Answer 5

a quotient rule problem so we use
\[(\frac{f}{g})'=\frac{gf'-g'f}{g^2}\] with
\[f(x)=x^2-2,f'(x)=2x,g(x)=\sqrt{x^2-1},g'(x)=\frac{x}{\sqrt{x^2-1}}\]

Answer 6

how did you get down to that simplfy version because my answer was 2x/(x^2-1)^(1/2)+(x^2-2)*-2x/(x^2-1)^(3/2)

Answer 7

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

Answer 8

now multiply top and bottom by
\[\sqrt{x^2-1}\] to clear the compound fraction

Answer 9

the denominator will be
\[(x^2-1)^{\frac{3}{2}}\]

Answer 10

the numerator will be
\[(x^2-1)\times 2x-(x^2-2)x\] and algebra should do the rest

Answer 11

what happen the square root for (x^2-1)?

Answer 12

in the denominator or numerator?

Answer 13

both

Answer 14

in the numerator you are multiplying
\[\sqrt{x^2-1}\times (\sqrt{x^2-1}\times 2x-(x^2-2)\times \frac{x}{\sqrt{x^2-1}})\]

Answer 15

the first term becomes
\[(x^2-1)\times 2x\]

Answer 16

and the second term is just
\[-(x^2-2)\times x\] by cancellation

Answer 17

in the denominator you have
\[(x^2-1)\times \sqrt{x^2-1}=(x^2-1)^{\frac{3}{2}}\] by the laws of exponents

Answer 18

wait so you would treat the term separately like for example multiplying square-root of (x^2-1) by 2X on the numerator

Answer 19

Oops I meant wouldn't not would

Answer 20

lets go slow

Answer 21

this is what we get just by applying the quotient rule directly
\[\frac{\sqrt{x^2-1}\times 2x-(x^2-2)\times \frac{x}{\sqrt{x^2-1}}}{x^2-1}\]

Answer 22

but this is a compound fraction, so we have to clean it up somehow

Answer 23

it is really this
\[\frac{\sqrt{x^2-1}\times 2x- \frac{x(x^2-2)}{\sqrt{x^2-1}}}{x^2-1}\]

Answer 24

ok

Answer 25

and we can clean up the compound fraction by multiplying top and bottom by
\[\sqrt{x^2-1}\]

Answer 26

that part where I'm still wondering why you don't multiply it by 2X

Answer 27

for the denominator you get
\[(x^2-1)\times \sqrt{x^2-1}=(x^2-1)\times (x^2-1)^{\frac{1}{2}}=(x^2-1)^{\frac{3}{2}}\]

Answer 28

multiply what by 2x?

Answer 29

square of (x^2-1) on the denominator

Answer 30

square root

Answer 31

no need

Answer 32

the only part that is is a compound fraction is in the numerator. it is like having
\[\frac{a+\frac{b}{c}}{d}\]

Answer 33

then you would multiply top and bottom by c to clear the fraction

Answer 34

yes, but isn't the a in the formula your using are square root (x^2-1) and 2x on the numerator, so you need to multiply both by square root of (x^2-1) on denominator

Answer 35

you only multiply the square root (x^2-1) on numerator by the square root (x^2-1) from the denominator, but forgot about the 2X

Answer 36

no i only need to multiply top and bottom by c, not by a

Answer 37

maybe i am confusing you. this is what i think you need
\[\frac{\sqrt{x^2-1}}{\sqrt{x^2-1}}\frac{\sqrt{x^2-1}\times 2x-(x^2-2)\times \frac{x}{\sqrt{x^2-1}}}{x^2-1}\]

Answer 38

this will clear the fraction for you

Answer 39

ok should the numerator have x^3 for final simply solution

Answer 40

lets see

Answer 41

the numerator should be
\[(x^2-1)\times 2x-(x^2-2)\times x\]
\[2x^3-2x-x^3+2x\] yes i get
\[x^3\]

Answer 42

ok so your final answer is x^3/x^2-1

Answer 43

Thanks!

Answer 44

wel actually it would be
\[\frac{x^3}{(x^2-1)^\frac{3}{2}}\]

Answer 45

yeah I just forgot to type it in, thanks again