Question

What is the derivative of x+(4/x) using the limit process?

Answer 1

\[f'(x)=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h} \\ =\lim_{h \rightarrow 0}\frac{[x+h+\frac{4}{x+h}]-[x+\frac{4}{x}]}{h} \\ =\lim_{h \rightarrow 0}\frac{1}{h}[x+h+\frac{4}{x+h}-x-\frac{4}{x}]\]
We want to try to cancel that lonely h on the bottom (the one outside the brackets)

Answer 2

first of all do you see anything that cancels inside the [ ]

Answer 3

The x cancels out

Answer 4

what i ended up with was \[(x(x^2+xh+h^2+4)-(x+h)(x^2+4)) / (x+h)(x)\]

Answer 5

but so far ive gotten stuck

Answer 6

so you tried to combine the fractions inside the brackets after x-x=0

Answer 7

yes

Answer 8

\[\frac{1}{h}[\frac{h(x)(x+h)+4(x)-4(x+h)}{x(x+h)}]\]

Answer 9

so on top there we have hx(x+h)+4x-4x-4h

Answer 10

you should see something else that cancels

Answer 11

the 4xs, but this is just the continuation of crossing out the original Xs right? cause I think I just did it in a way that made the problem a whole lot more complicated for me

Answer 12

Yeah the 4x's cancel
I can't really tell how you got your fraction up there...
Do you understand how I got mine fraction?

Answer 13

No I'm not really sure... I think I have an idea

Answer 14

\[f'(x)=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h} \\ =\lim_{h \rightarrow 0}\frac{[x+h+\frac{4}{x+h}]-[x+\frac{4}{x}]}{h} \\ =\lim_{h \rightarrow 0}\frac{1}{h}[x+h+\frac{4}{x+h}-x-\frac{4}{x}] \\ =\lim_{h \rightarrow 0}\frac{1}{h}[\cancel{x} +h+\frac{4}{x+h}-\cancel{x}-\frac{4}{x}] \\ =\lim_{h \rightarrow 0}\frac{1}{h}[h+\frac{4}{x+h}-\frac{4}{x}] \\ \\ \text{ lcm of the bottoms is x(x+h) } \\ =\lim_{h \rightarrow 0}\frac{1}{h}[h \frac{x(x+h)}{x(x+h)}+\frac{4}{x+h}\frac{x}{x}-\frac{4}{x} \frac{x+h}{x+h}] \\ \lim_{h \rightarrow 0}\frac{1}{h}[ \frac{hx(x+h)+4x-4(x+h)}{x(x+h)}]\]

Answer 15

That is how I combined the fractions inside the []

Answer 16

now after canceling those 4x's you will that all that remains are terms in the numerator that have a common factor h so you can factor that h on top and cancel it with the bottom lonely bottom h outside the [ ]

Answer 17

Okay I just figured it out at the same time you sent me that!

Answer 18

one you get rid of that factor h there that is on bottom (with the top factor h)
you will be able to replace remaining h's with 0's and find f'(x)

Answer 19

Wait I'm sorry Im confused so I have \[hx(x+h)-4h / h\] and I am factoring out ...?

Answer 20

\[f'(x)=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h} \\ =\lim_{h \rightarrow 0}\frac{[x+h+\frac{4}{x+h}]-[x+\frac{4}{x}]}{h} \\ =\lim_{h \rightarrow 0}\frac{1}{h}[x+h+\frac{4}{x+h}-x-\frac{4}{x}] \\ =\lim_{h \rightarrow 0}\frac{1}{h}[\cancel{x} +h+\frac{4}{x+h}-\cancel{x}-\frac{4}{x}] \\ =\lim_{h \rightarrow 0}\frac{1}{h}[h+\frac{4}{x+h}-\frac{4}{x}] \\ \\ \text{ lcm of the bottoms is x(x+h) } \\ =\lim_{h \rightarrow 0}\frac{1}{h}[h \frac{x(x+h)}{x(x+h)}+\frac{4}{x+h}\frac{x}{x}-\frac{4}{x} \frac{x+h}{x+h}] \\ =\lim_{h \rightarrow 0}\frac{1}{h}[ \frac{hx(x+h)+4x-4(x+h)}{x(x+h)}] \\ =\lim_{h \rightarrow 0}\frac{1}{h}[\frac{xh(x+h)-4h}{x(x+h)}] \\ =\lim_{h \rightarrow 0}\frac{h}{h} [\frac{x(x+h)-4}{x(x+h)}]\]

Answer 21

now remember h/h=1
so now you have gotten rid of the thingy that would make the bottom 0
so replace remaining h's with 0

Answer 22

Okay so I'm just left with \[\left( x^2-4\right)/x^2\]

Answer 23

looks good

Answer 24

why wouldn't I just cancel out the x(x+h)

Answer 25

\[f'(x)=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h} \\ =\lim_{h \rightarrow 0}\frac{[x+h+\frac{4}{x+h}]-[x+\frac{4}{x}]}{h} \\ =\lim_{h \rightarrow 0}\frac{1}{h}[x+h+\frac{4}{x+h}-x-\frac{4}{x}] \\ =\lim_{h \rightarrow 0}\frac{1}{h}[\cancel{x} +h+\frac{4}{x+h}-\cancel{x}-\frac{4}{x}] \\ =\lim_{h \rightarrow 0}\frac{1}{h}[h+\frac{4}{x+h}-\frac{4}{x}] \\ \\ \text{ lcm of the bottoms is x(x+h) } \\ =\lim_{h \rightarrow 0}\frac{1}{h}[h \frac{x(x+h)}{x(x+h)}+\frac{4}{x+h}\frac{x}{x}-\frac{4}{x} \frac{x+h}{x+h}] \\ =\lim_{h \rightarrow 0}\frac{1}{h}[ \frac{hx(x+h)+4x-4(x+h)}{x(x+h)}] \\ =\lim_{h \rightarrow 0}\frac{1}{h}[\frac{xh(x+h)-4h}{x(x+h)}] \\ =\lim_{h \rightarrow 0}\frac{h}{h} [\frac{x(x+h)-4}{x(x+h)}] \\ \text{ you could write this as } \\ =\lim_{h \rightarrow 0}\frac{h}{h} [1-\frac{4}{x(x+h)}]\]

Answer 26

That will still lead to the same answer...

Answer 27

ooooh okay thankss!

Answer 28

\[\frac{x^2-4}{x^2}=\frac{x^2}{x^2}-\frac{4}{x^2}=1-\frac{4}{x^2}\]

Answer 29

hahaha Idk why I didn't see that earlier

Answer 30

thank you so much, this was a tough one