Question

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Answer 1

do you now the definition of derivative as a limit ?

Answer 2

not off hand no, i missed the lecture on this specifically

Answer 3

\(\large f'(x) = \lim \limits_{h \to 0}\frac{ f(x+h)-f(x) }{ h }\)

Answer 4

so, since f(x+y) = f(x) f(y)
what will be f(x+h) =...?

Answer 5

yeah confused there since its not given.. if fx+y is f(x)*f(y) doesnt that leave the second part of the equation for f(x+h)?

Answer 6

uhhh..?
f(x+y) = f(x) f(y)
put y= h in this
what do u get ?

Answer 7

ohhhhh f(x)*f(h) sorry

Answer 8

yesss

Answer 9

our numerator = f(x+h) - f(x) = f(x) f(h) - f(x)
what gets factored out ?

Answer 10

x^2

Answer 11

i don't see x^2 anywhere :/

Answer 12

f(x)*f(h)-f(x) we are simplifying right?

Answer 13

yes

Answer 14

so cant we combine the two f(x)s?

Answer 15

a*b - a = a (b-1)
f(x)*f(h)-f(x) =...?

Answer 16

f(x)[f(h)-f(x)] then, wow i was off base

Answer 17

you mean
f(x) [f(h) - 1]
?

Answer 18

yes sorry

Answer 19

no problem, now lets wonder how do we get f(h) -1

Answer 20

we have
f(x) = 1+x g(x)
put x= h here, what do u get ?

Answer 21

1+h*g(x)

Answer 22

g(h)*

Answer 23

yes
so
f(h) -1 =... ?

Answer 24

h*g(h)?

Answer 25

correct!

Answer 26

plug this in our limit formula
what gets cancelled ?

Answer 27

f′(x)=limh→0 f(x+h)−f(x)h...
so i'm plugging in h*g(h) for all the h's here?

Answer 28

let me show you..

Answer 29

\(\large f'(x) = \lim \limits_{h \to 0}\frac{ f(x+h)-f(x) }{ h } \\ \large \large f'(x) = \lim \limits_{h \to 0}\frac{ f(x)(f(h)-1) }{ h } \\ \large \large f'(x) = \lim \limits_{h \to 0}\frac{ f(x)[h g(h)] }{ h }\)
can you now tell what gets cancelled ?

Answer 30

youre left with f(x)[g(h)]?

Answer 31

correct!
now since h is the variable here,
as h-> 0
f(x) is the constant and can be pulled out of the limit
\(\large f'(x) = \lim \limits_{h \to 0}\frac{ f(x+h)-f(x) }{ h } \\ \large \large f'(x) = \lim \limits_{h \to 0}\frac{ f(x)(f(h)-1) }{ h } \\ \large \large f'(x) = \lim \limits_{h \to 0}\frac{ f(x)[h g(h)] }{ h } \\ \large f'(x) = f(x) \lim \limits_{h \to 0 }g(h)\)
got this step clearly ?

Answer 32

ok

Answer 33

so then we plug that back in?

Answer 34

limit x(approaches 0) g(x)=1.
put x= h here, what do u get ?

Answer 35

h(app. 0)g(h)=1

Answer 36

so what is the value of the enlarged green part ?
\( \large f'(x) = f(x) \huge \color {green }{\lim \limits_{h \to 0 }g(h)}\)

Answer 37

and what remains finally ?

Answer 38

1..?

Answer 39

yes

Answer 40

so that's it since we showed it was equal to 1 and the given for f(x) was 1, thus f'(x)= f(x)?

Answer 41

yes,
that limit = 1
so
f'(x) = f(x)
which was exactly what we needed to prove :)

Answer 42

alright cool i will look over this and write it down, thanks a lot

Answer 43

welcome ^_^