Question

Anyone know how to derive using formal definition?

Answer 1

yes, it usually involves a bunch of algebra first
what example do you have?

Answer 2

\[f(x)=(3x^2)-5x\]

Answer 3

could either use
\[\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}\] or
\[\lim_{x\to a}\frac{f(x)-f(a)}{x-a}\] depending on how you are looking at it

Answer 4

prefer the second

Answer 5

\[f(x)=3x^2-5x\] like that?

Answer 6

yes

Answer 7

lets take it slow, because almost all of it is algebra

Answer 8

\[f(x)=3x^2-5x\] we need also \[f(x+h)=3(x+h)^2-5(x+h)\] now it is a matter of the distributive law repeatedly

Answer 9

\[f(x+h)=3(x+h)^2-5(x+h)=3(x^2+2xh+h^2)-5x-5h\] is the first step, then continue to distribute and get
\[f(x+h)=3x^2+6xh+3x^2-5x-5h\]

Answer 10

now for \(f(x+h)-f(x)\) you get \[3x^2+6xh+3x^2-5x-5h-(3x^2-5x)\] distribute again and get\[3x^2+6xh+3x^2-5x-5h-3x^2+5x\]

Answer 11

as usual everything without an \(h\) in it goes away, leaving only
\[f(x+h)-f(x)=6xh-5h+3h^2\]

Answer 12

damn made a typo
should be
\[f(x+h)=3x^2+6xh+3\color{red}h^2-5x-5h\]

Answer 13

this line \[f(x+h)-f(x)=6xh-5h+3h^2\] is correct though.
now
\[\frac{f(x+h)-f(x)}{h}\] cancel an \(h\) everywhere and get
\[6x-5+3h\]

Answer 14

and finally "take the limit as \(h\to 0\) by replacing \(h\) by \(0\) to finish with
\[f'(x)=6x-5\]

Answer 15

as you can see, almost all of it is algebra
the "limit" business only comes in at the very end

Answer 16

f'(0)=6*0-5
f'(0)=0-5
f'(0)=-5

Answer 17

if you are looking for \(f'(0)\) then it is \(-5\) right

Answer 18

you can also find it by computing
\[f'(0)=\lim_{x\to 0}\frac{f(x)-f(0)}{x-0}\] if you like
want to try it that way?

Answer 19

oh i thought you said we had to

Answer 20

i found \(f'(x)\) using the definition
we can also find \(f'(0)\) using the second definition if you would like to

Answer 21

i did
\[f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}\] but we can also find the number
\[f'(0)=\lim_{x\to a}\frac{f(x)-f(0)}{x-0}\]
that takes way less algebra

Answer 22

thank you