Find Derivative of f(x)= 1/(2x-4) Using Limit Definition

[Precal_Chris]

Homework Statement

Use the limit definition to find the derivative of the function.
f(x)= 1/(2x-4)

Homework Equations

f(x+h)-f(x)/h

The Attempt at a Solution

ok so first i plugged it all in..

(1/(2(x+h)-4))-(1/(2x-4)) / h

from here i was going to do fly-by with the two top fractions but for some reason nothing cancels out...
so please help this is due for me by monday! lol
(any help appreciated)

 

[Snazzy]

Expand it, find a common denominator, move on from there. You're on the right track.

 

[Precal_Chris]

ok so what about this?

for the top fractions if you do fly by you get:

(2x+2h-4 - 2x+4)/(2x+2h-4)(2x+4) and then of course all of that divided by h...

then you have just 2h on the numerater becuz everything else cancels out...
so:

(2h)/ (2x+2h-4)(2x+4) all divided by h...

so then you can do the following...

2h/(2x+2h-4)(2x+4) • 1/h (reciprical)

by doing that you can cancel out the h's

2/(2x+2h-4)(2x+4) •1...

thats as far as i can go...is this right? if so what do i do next please

 

[Tedjn]

That is almost right. Check your denominator again, it should not be (2x+2h-4)(2x+4). The rest of your algebra is correct, and you did well to cancel out the h. Now, recall the definition of a derivative. What should the limit be?

 

[Precal_Chris]

i have no idea because when she was teaching this portion i was making up a quiz :/ so that's why I am so lost..
and should the denominator be (2x+2h-4)(2x-4) ? becuz i thought the whole part with
f(x+h) - f(x) i thought that changed the sign of each part? due to the subtraction sign

 

[Tedjn]

Yes, the denominator should be (2x+2h-4)(2x-4). Even though you are subtracting two fractions, you still need to find a common denominator, which, in this case, is a product of the two denominators.

I've taken a quick glance at some of your other threads, and you seem to be getting good advice there. This is the same thing. Apply the limit definition of the derivative that you were given, replacing the generic f(x) with your specific function. Then, you need to simplify to a point where you can replace the variable you are taking to a limit with the limiting value and still get a meaningful answer (something other than 0/0), and you will have the derivative.

 

[Precal_Chris]

yeah i just don't know what the limit should be for this specific problem.. because it didnt give me a point

 

[Snazzy]

You don't need a point. The limit as h->0 is for any point, x, on the derivative of the function. It will be in terms of x.

 

[Precal_Chris]

so 2/(2(0) + 2h +4)(2(0) -4)
2/(2h+4)(-4)
2/(-8h -16)

is that right? can i just leave the h there like that?

 

[sutupidmath]

\lim_{x\rightarrow a}\frac{\frac{1}{2x-4}-\frac{1}{2a-4}}{x-a}=\lim_{x\rightarrow a}\frac{2a-4-2x+4}{(2x-4)(2a-4)(x-a)}=\lim_{x\rightarrow a}\frac{-2(x-a)}{(2x-4)(2a-4)(x-a)}=-2\lim_{x\rightarrow a}\frac{1}{(2x-4)(2a-4)}=\frac{-2}{(2a-4)^{2}}=f'(a), \forall a



P.S. YOu have another post with the exact same question.

 

[Snazzy]

so 2/(2(0) + 2h +4)(2(0) -4)
2/(2h+4)(-4)
2/(-8h -16)

is that right? can i just leave the h there like that?

It's as h->0, not as x->0. Also, 2/ ((2x+2h+4)(2x-4)) is incorrect.

 

[Precal_Chris]

oops lol
dont i feel dumb...

ok so its:
2/(2x+2(0)-4)(2x-4)
2/(2x-4)(2x-4)
2/4x^2-16x+16
and that would be my answer?

 

[Snazzy]

\lim_{h\rightarrow 0}\frac{\frac{1}{2x+2h-4}-\frac{1}{2x-4}}{h}

Do it again starting from here, mate. Your answer's not right.

 

[Astronuc]

Precal_Chris,

Do as Snazzy suggested and this time write out the numerator paying attention to the order.

 

[Precal_Chris]

oh ok..so wait when i do fly by... is my numerator supposed to be:

2x+2h-4-2x-4?

which would ultimately make it..

2h-8 then the h's cancel so i have

-6 in the numerator?

 

[HallsofIvy]

No, it is (2x+2h-4)-(2x-4)

Be careful with your signs!

 

[Precal_Chris]

well then wouldn't my first answer be right?
cuz
2x+2h-4 -2x +4 would cancel the 2x and the 4's out
so id be left with 2h..
then the h would cancel
and id have
2/(2x+2h-4)(2x-4)

and then after putting in the 0's for the h..
id have

2/(2x-4)(2x-4)
so that would be
2/(2x-4)^2
or
2/(4x^2-16x+16)
right?

 

[sutupidmath]

well then wouldn't my first answer be right?
cuz
2x+2h-4 -2x +4 would cancel the 2x and the 4's out
so id be left with 2h..
then the h would cancel
and id have
2/(2x+2h-4)(2x-4)

and then after putting in the 0's for the h..
id have

2/(2x-4)(2x-4)
so that would be
2/(2x-4)^2
or
2/(4x^2-16x+16)
right?

You are still short of a minus sign, check it with post #10.

 

[Precal_Chris]

in post #10 i didnt understand nething on it..it had a different equation than what i was working with
but i already had other people tell me that it was right on this ...so when i brought it up later it should still be right?

 

[Precal_Chris]

oh wait i understand now...
when i did flyby i did it in the incorrect order...making the numerator wrong...
instead of:

(2x+2h-4)-(2x-4)
it should have been

(2x-4)-(2x+2h-4)

then after losing the x's and the 4's i'd have:

-2h/(2x+2h-4)(2x-4) • 1/h

then i get rid of the h

so:
-2/(2x+2h-4)(2x-4)

then plug in 0 for h

-2/(2x+0-4)(2x-4)

-2/(2x-4)^2

is that my final answer?

 

[sutupidmath]

in post #10 i didnt understand nething on it..it had a different equation than what i was working with
but i already had other people tell me that it was right on this ...so when i brought it up later it should still be right?

Yes, it has to be right, if you do the calculations right.

There are two ways of expressing the def. of the derivative of a function.
The first one is in terms of the points, that is

f'(a)=\lim_{x\rightarrow a}\frac{f(x)-f(a)}{x-a}

and the other is in terms of the distance between two points, h

f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}

YOu can choose whichever one you like. Usually you should choose the one that gets you faster to the answer, and obviously using the first one f'(a), as i did in post #10 gets you way faster to the answer than the second one.

Use whichever one you like.

THese two are the same thing since h=x-a, \ \ \ so \ \\ when \ \ \ \ x--->a,h-->0 and

h=x-a=>x=h+a so f(x)=f(h+a), and x-a becomes h+a-a=h.

 

[sutupidmath]

-2/(2x-4)^2

is that my final answer?

Yes!

 

[Precal_Chris]

wow that's great...
thanks guys now i just have that last one to answer and I am set...