Derivative of 1/2sin2x

Question

psymon · Answer

Well, we need to move this into the numerator so we can take the derivative of properly. So let's first rewrite it like this:
$$\frac{ (\sin2x)^{-1} }{ 2 }$$So now this just becomes a chain rule. Would you know how to do chain-rule from here?

anonymous · Answer

If you chain rule sin2x dosnt it become 0?

psymon · Answer

Nope, not at all. So chain rule is when you take the derivative of each layer (or inner function) and then multiply those results. In this problem we have 3 layers/functions
Note: Trig functions always count as 2 layers, the function and the angle
So these are our 3 layers:
$$(----)^{-1}  $$
$$\sin(--)$$
$$2x $$
I just leave the blanks so I can emphasize what the layers we are taking the derivative of actually are. So let's start with the first one. using the power rule, could you simplify the first one?

anonymous · Answer

$$-(----)^{-1+1}$$

psymon · Answer

-1 remember.

psymon · Answer

So right, we would have:
$$-(---)^{-2} $$
So now we look at the sin part. what would be the derivative of that second layer?

anonymous · Answer

$$-(\sin2x)^{-2}(2\cos2x)$$

psymon · Answer

Lol, you were getting the idea, cool xD Exactly, you take the derivative of all 3 layers multiply them, and plug back in what was inside each layer, good job. So now we can't forget that we had a 2 in the denominator. So that 2 on top and bottom would cancel and we'd have:
$$-\frac{ \cos2x }{ \sin ^{2}2x }  $$ Because I moved it back into the denominator to make it positive. So that would be your answer : )

anonymous · Answer

Ok, thanks I was thinking you had to add one to the power

psymon · Answer

thats integration :3

anonymous · Answer

Yea, Ive been doing more integration than derivatives lately

psymon · Answer

Yeah, when you first doing it you can definitely gets things backwards, especially with the trig stuff. Going from sin to cos or cos to sin, etc, lol.