Question

let f(x)=5/(x-6)
according to the definition of a derivative,
f'(x)=lim as t approaches x ____________________

Answer 1

You probably need to re-phrase the question.

Where does "t" come in?

Answer 2

i actually have no clue. the question just says use variables x and t and that's all it gives! thanks for trying though!

Answer 3

Here's the question! please help!

Answer 4

f'(x)=\frac{-5}{x-6}^2

Answer 5

maybe there t->0
so, f'(x)=lim(t->0) (f(x+t)-f(x))/t

Answer 6

\[f'(x)= \frac{-5}{(x-6)^2}\]

Answer 7

yea, i got like that too
but i think @tenistaego want u show get this

Answer 8

well, i want to show how to get it!
lim(t->0) (f(x+t)-f(x))/t
= lim(t->0) (5/(x+t-6)-(5/(x-6))/t
= lim(t->0) (5x-30-5x-5h+30)/(h*(x+h-6)*(x-6))
= lim(t->0) -5h/(h*(x+h-6)*(x-6))
= lim(t->0) -5/(x+h-6)*(x-6))
just put h=0
u will get the answer like on!

Answer 9

am i right, @hartnn ? hehe..

Answer 10

heyy! thanks for the answer @RadEn! just wondering if you could look at this picture and tell me what the first box is? thanks!

Answer 11

yea, i see
but actually if it is a definition of derivative y=f(x), so
y' = f'(x) = lim (t->0) (f(x+t)-f(x))/t
i've never heard lim (t->x), sometimes there is told me
f'(x) = lim (h->0) (f(x+h)-f(x))/h

Answer 12

ahh it says its not correct! haha its ok thanks for trying though RadEn

Answer 13

you're welcome :)