1st and 2nd derivative of :

2sin2x??

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.

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

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.

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