Check out some similar questions!

kookai18 · Aug 30, 2009, 06:02 PM

1st and 2nd derivative of :

2sin2x??

ebaines · Aug 31, 2009, 10:51 AM

Not knowing how far along you are...

You should memorize the following facts: the derivative of sin(x) is cos(x), and the derivative of cos(x) is -sin(x).

You should also know that the derivative of f(g(x)) is f'(g(x))*g'(x).

With this you should be able to get the answer. Post back if you're still having difficulty.

kookai18 · Aug 31, 2009, 05:04 PM

thanks for your reply ebaines... I know those derivitives by heart but I'm having problems deriving this particular one (2sin2x)... if you could help me id really like that

ebaines · Sep 1, 2009, 05:58 AM

What you have is f() = sin() and g(x) = 2x. So
$ \frac {df(g(x))} {dx} = f'(g(x))*g'(x). $
Since g(x) = 2x, then g'(x) = 2. So you get:

$ \frac {d(sin(2x))} {dx} = cos(2x)*2 $

Now, don't forget that initial constant of 2:

$ \frac {d(2*sin(2x))} {dx} = 2*cos(2x)*2 = 4 cos(2x) $

Now, can you get the second derivative? You use the same process.

kookai18 · Sep 1, 2009, 07:32 PM

Thank u , I have tried it before you helped me and I got the same answer :) thanks for your help:D