Question

From first principles, find the derivative of y=-2sin3x

Answer 1

f'(x) =lim (h tends zer0) (f(x+h)-f(x))/h
Use this formula find derivative

Answer 2

okay, i used the f(x) - f(a) to start off with but i will try this now

Answer 3

good luck

Answer 4

you need the "addition angle" formula for this
plus you need to know a couple limits

Answer 5

thanks !

Answer 6

ignore the \(-2\) it has no effect on the limit, you can put it in at the end
then compute
\[\frac{\sin(3(x+h))-\sin(3x)}{h}\] or
\[\frac{\sin(3x+3h)-\sin(3x)}{h}\] using the "addition angle formula to go to
\[\frac{\sin(3x)\cos(3h)+\sin(3h)\cos(3x)-\sin(3x)}{h}\]

Answer 7

then break apart in to two pieces
if you have a proof in your book that the derivative of sine is cosine, you can mimic that, with the only difference that this time the input is \(3x\) and not \(x\)
it will work pretty much the same

Answer 8

okay I'll have a look in my textbook now. thanks for you help so far!

Answer 9

yw

Answer 10

if I have \[\lim_{\theta \rightarrow 0} \frac{ \sin3h }{ h } = 1\]

does it actually equal one? Im not sure what to do with the 3 bit...

Answer 11

not it does not

Answer 12

\[\lim_{h\to 0}\frac{\sin(3h)}{h}=3\] for the following reason

Answer 13

oooOH i see

Answer 14

\[\lim_{h\to 0}\frac{\sin(3h)}{h}=3\lim_{h\to 0}\frac{\sin(3h)}{3h}=3\times 1=3\]

Answer 15

cool! thanks for explaining :)

Answer 16

you did expect a three right? because you know the answer already

Answer 17

yw

Answer 18

im still going, but yes i do know the answer

Answer 19

YAY i got it! thanks for your help – it is greatly appreciated! love your work

Answer 20

you are quite welcome
i appreciate the compliment

Answer 21

:)

Answer 22

i got a question, if i've got \[\frac{ \cos 3h -1 }{ h }\] then as h-->0 why does cos3h--> 0?

Answer 23

this is what i am referring to: