Question

Can someone help me find the derivative of the f(x): (8)/(sqrt x-2), using the difference quotient

Answer 1

Let f(x)=8, and g(x)=(sqrt(x-2)
\[(f'g-g'f)/g^2\]
f(x)=8. f'(x)=0
\[g(x)=\sqrt{x-2}\]
\[d/dx (f(x)/g(x)) = \frac{g'(x)f}{g^2}\]

Answer 2

can someone help me using the difference quoteint: (f(x+h)-f(x))/h

Answer 3

when you say "difference quotient" you mean the definition yes?

Answer 4

yes

Answer 5

\[lim_{h->0}\frac{f(x+h)-f(x)}{h}\]

Answer 6

\[f(x)=\frac{8}{\sqrt{x-2}}\] yes?

Answer 7

yes

Answer 8

ok for ease of writing i am going to skip writing the limit each time. we will take the limit at the end.

Answer 9

ok

Answer 10

and also just remember that we are going to have to divide by h at some point. are main job is to simplify
\[f(x+h)-f(x)=\frac{8}{\sqrt{x+h-2}}-\frac{8}{\sqrt{x-2}}\]

Answer 11

ok

Answer 12

so we just subtract and get
\[\frac{8\sqrt{x-2}-8\sqrt{x+h-2}}{\sqrt{x+h-2}\times\sqrt{x-2}}\]

Answer 13

now we can pull the 8 right in front of the limit, but we could have done that at the beginning. the gimmick now is to multiply top and bottom by the conjugate of the numerator, namely
\[\sqrt{x-2}+\sqrt{x-2+h}\]

Answer 14

alright

Answer 15

when you do that , your numerator will be
\[x-2-(x+h-2)=-h\]

Answer 16

which is nice because this is the h that will cancel with the h in the denominator

Answer 17

you denominator will be just what you get, and ugly mess namely
\[h\times\sqrt{x+h-2}\times \sqrt{x-2}(\sqrt{x-2}+\sqrt{x+h-2})\]

Answer 18

ok

Answer 19

but now you can cancel the h from the numerator and denominator. you numerator will just be -1 (or -8 when you bring the 8 back) and the denominator will be what i wrote about without the in it namely
\[\sqrt{x+h-2}\times \sqrt{x-2}(\sqrt{x-2}+\sqrt{x+h-2})\]

Answer 20

how do we clean up the denominator

Answer 21

well now you can take the limit as h->0 because you will not get 0/0 if you do. replace h by zero now

Answer 22

in other words take
\[lim_{x->0}\frac{-8}{\sqrt{x+h-2}\times \sqrt{x-2}(\sqrt{x-2}+\sqrt{x+h-2}})\]

Answer 23

no more zero over zero. we solved that problem when we canceled the h in the denominator with the h in the numerator. so now wherever you see an "h" replace it by 0

Answer 24

The 8 that we took out now gets placed back in the numerator

Answer 25

actually look smart and put it right outside the limit sign

Answer 26

that really should be step 1

Answer 27

so what would be the derivative, if you dont mind

Answer 28

but we do put the 8 back into the num inorder to get the derivative right?

Answer 29

the first line should look like
\[8 lim_{h->0}\frac{1}{h} \times (\frac{1}{\sqrt{x+h-2}}-\frac{1}{\sqrt{x-2}})\]

Answer 30

take the 8 right out front. yes we put it at the end

Answer 31

so now your last line should look like
\[8\times \frac{-1}{\sqrt{x-2}\times \sqrt{x-2}(\sqrt{x-2}+\sqrt{x-2})}\]

Answer 32

i just replaced all h's in the denominator by 0

Answer 33

in the expression about 11 lines up

Answer 34

I got it thank you

Answer 35

now some algebra to get
\[\frac{-8}{(x-2)(2\sqrt{x-2})}=\frac{-4}{(x-2)\sqrt{x-2}}\]

Answer 36

I might be asking to much of you, but could you help me understand how to use the alternative formula for derivative, namely : f prime (x)= f(z)-f(x)/x-x

Answer 37

z-x on the denominator

Answer 38

perhaps of x^2-3x+4

Answer 39

you mean
\[lim_{z->x}\frac{f(z)-f(x)}{z-x}\]

Answer 40

yes

Answer 41

this is really just an exercise in algebra for your example. it should go quick. especially since we know the answer is
\[f'(x)=2x-3\]

Answer 42

trick of course will be to factor an (z-x) out of the numerator and cancel.

Answer 43

okay

Answer 44

we get
\[\frac{z^2-3z+4-(x^2-3x-4)}{z-x}\]
\[=\frac{z^2-x^2-3z+3x}{z-x}\]

Answer 45

okay

Answer 46

now look to factor. first term is diffeernce of two squares, second to obviously have common factor of 3

Answer 47

\[\frac{(z-x)(z+x)-3(z-x)}{z-x}\]

Answer 48

or
\[\frac{(z-x)(z+x-3)}{z-x}\]

Answer 49

cancel the z - x and get
\[z+x-3\] and now take the limit as z ->x by simply replacing the x by a z to get
\[2x-3\] and we knew

Answer 50

i meant of course "by simply replacing the z by an x"

Answer 51

in every one of these problems, you goal is to factor and cancel

Answer 52

okay, i was confused about what to do with that form

Answer 53

either factor out an "h" in the first form, or factor out a (z-x) in the second

Answer 54

thank you

Answer 55

they are identical with h replacing z - x

Answer 56

yw