Question

derivative question

Answer 1

\[f(x)=\frac{ \sqrt{7}-x }{ \sqrt{7}+x}\]

Answer 2

find f(x)'

Answer 3

do you know how to use the quotient rule?

Answer 4

yes \[\frac{ gf'-fg' }{ g^2 }\]

Answer 5

but its sqrt...!

Answer 6

x^(1/2)?

Answer 7

You differentiate a constant \(\sqrt{7}\), you get zero

Answer 8

yea don't worry about that you are taking the derivative in respect to x

Answer 9

\[\sqrt{7}\] is a constant so when you take the derivative of a constant it goes to 0

Answer 10

So i get zero????
I thought I had to do 7^1/2...

Answer 11

and the derivative of x is 1 so you get

\[\frac{ -1(\sqrt{7}+x)-1(\sqrt{7}-x) }{ (\sqrt{7}+x)^2 }\]

now can you simplify that and get f'

Answer 12

1? how?

Answer 13

nope sqrt(7) is a constant as long as there is no variable next to it the derivative will always be 0

Answer 14

\[\frac{ -1(0+x)-1(0-x) }{ ((\sqrt(7)+x)^2 }\]
??

Answer 15

ok first distribute the 1 to each component in the parenths, you will get

\[\frac{ -\sqrt{7}-x-\sqrt{7}+x }{ (x+\sqrt{7})^2 }\]

Answer 16

you can see that the x's will cancel out. and you can see that its \[-\sqrt{7}-\sqrt{7}=2\sqrt{7}\]

Answer 17

leave the bottom the same and then you will have

\[\frac{ -2\sqrt{7} }{ (x+\sqrt{7})^2 }\]

Answer 18

I see!! whats next step?

Answer 19

lol that is the answer

Answer 20

Oh, if the question asks f(5)'.
then i just need to insert 5 into the x ?

Answer 21

yea just plug in x

Answer 22

and solve because everything is now a constant

Answer 23

Thank you very much

Answer 24

You welcome