Question

find the derivative of f(x)=cosx by first principle of derivatives

Answer 1

First principle?
\[\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}\]
Plug in x+h then use the identity
cos(a+b)=cos(a)cos(b)-sin(a)sin(b)
Then rewrite and reduce to get -sin(x)

Answer 2

i tried but i didnt get it can you show me :(

Answer 3

Let me see, I've never actually worked this out. H/o.

Answer 4

i got (cosxcosh-sinxsinh-cosx)/h

Answer 5

sigh...i need to go study my indentities...

Answer 6

Exactly. From here group it:
\[\lim_{h \rightarrow 0}\cos(x)\frac{\cos(h)-1}{h}-\sin(x)\frac{\sin(h)}{h}\]
Working left to right the first limit you have:
\[\cos(x)\lim_{h \rightarrow 0}\frac{\cos(h)-1}{h}\]
gives you zero/zero. Use l'hospital's rule to evaluate it.
\[\cos(x)\lim_{h \rightarrow 0}-\sin(h)=0\]
Then the second limit you have:
\[-\sin(x)\lim_{h \rightarrow 0}\frac{\sin(h)}{h}=-\sin(x)\]
So you have 0-sin(x)=-sin(x). Your answer.

Answer 7

i never learned i'hospital's rule...

Answer 8

Without that, I don't know how to do it :/

Answer 9

oh okay lol........thanks so much again malevolence...i mean i may have earned it but not that term...

Answer 10

You can probably find it if you google it D:

Answer 11

okay thanks!