someone good with calc plz plz plzzzz help!!

[Finewth69]

ok so my calc teacher is maaad chill she rly cares about her students and wants us all to do well in this class she knows that its the hardest calc class in the school and she wants to give us all high chances to get an A
sooo
she offered us extra credits, 3% on any test, if we give this to her:
a proof of why the derivative of cos(x)=-sin(x)
yes i googled it BUT since I never took calc before, I went from pre calc straight to AP calc, i dont know the chain rule and like the other pre calcies im not allowed to use it, so i need to give a proof without the chain rule so if anyone can tell me how to prove it it wud b muchhh appreciated. i didnt try it yet but i think i know how to get it, but just in case can someone do it here, or tell me that if if i say cos(x)=sin(x+ pi/2) and get the derivative of that, wud i b going in da rite direction?

[Crazyme99]

woola what grade u in????

[ecstasy]

well, i know that if you plot the derivative of cos u get sin and vice versa... its hard to explain, but wenever the slope of the cos graph is zero, then it is an X intercept for the derivative, if you r using a ti 83+ graphing calculator, u can type into the graph screen
y1=cosx
y2=Nderive(y1,x,x)
(to get nderive, type "math" > "8")

and it will graph cos and its derivative function, sin
(to find y1, press "vars" > "yvars" > "y1"

but, im really not sure how to explain it in algebraic form but graphically yes, i actually just took a test where i had to graph the derivative function of it, like
f(x)=x^3 f'(x)=3x^2
where f' is its derivative function, so
f(x)=cosx f'(x)=? im not sure how to use the power rule on that :S
but i do know that wen the graph hits a high point or a low point and the slope is zero, then that is a intercept of the same y value on the x-axis, then u just connect the dots logically based on the slope of the curves.

i know this isnt much help but its all i can do :S

[raulperez]

if you have the limit definition of the derivative you can do it as follows:

Some definitions:
lim (h->x) of <something> means the limit of something as h approaches x
d_<something> means the derivative of something
from algebra/trig we know that cos(a+b) = cos(a)cos(b)-sin(a)sin(b)

use (lim (h->0) of sin(h)/h)=1 it's an identity, they shouldn't need proof of that
also (lim (h->0) of (cos(h)-1)/h)=0 also an identity
then apply the limit definition of a derivative
d_cos(x) = lim (h->0) of (cos(h+x)-cos(x))/h //from hereout i'll shorten "lim (h->o)" to just "lim"
= lim (cos(h)cos(x)-sin(h)sin(x)-cos(x))/h
=lim (cos(h)cos(x)-cos(x))/h - lim (sin(h)sin(x))/h
=lim (cos(x)(cos(h)-1))/h - lim (sin(h)sin(x)/h
substitute in the two identities defined earlier
=(cos(x)*0)-(sin(x)*1)
=-sin(x)

that take care of it?
if you need any clarifications just shoot me a msg at rageagainstlowe@hotmail.com

[farnagain]

proofs f*cking suck o.o if you need logirithim help, let me know xD

[Finewth69]

yup dat was it raul thanks man
i figured it out during physics lol i dun know y i found it hard but i did exactly what you did

@BHE
im 15 n in 11th grade

@imrealez
that was by far the most useless post on this forums man i have to say

[Crazyme99]

one grade above me

[ecstasy]

well tyvm