What is the 50th derivative of F(x)=Cos(2x)?

[Alem2000]

The qeustion was what is the 50th dy/dx of F(x)=Cos(2x)
this is what i did
1st------dy/dx=-2Sin(2x)
2nd------dy/dx=-4Cos(2x)
3rd-------dy/dx=8Sin(2x)
4th-------dy/dx=16Cos(2x)
5th-------dy/dx=-32Sin(2x)...now that it reapeated itself its time to figure out the 50th. that's where I am confused. My problem is with the negative signs. how can i catch that patteren...

 

[AKG]

If you think about it, it should be pretty clear. Differentiating cosine gives a negative, differentiating sine gives none. And you differentiate a cosine every other time.
So, you start with
-sin. Now, differentiating sin gives no negative, so nothing changes
-cos. Cos changes things, so
+sin
+cos
-sin
-cos
+
+
-
-
+
+
.
.
.

Got it?

 

[Alem2000]

hmm...yeah i see. Thanks!

 

[nta]

change to cosine function

I think you should use this way to solve your problem:
1th dx/dy = -2sin2x=+2cos(2x+pi/2), so we obtain
2th dx/dy = +4cos(2x+pi)
3th dx/dy = +8cos(2x+3pi/2)
Continueing, we obtain:
i th dx/dy=+2^icos(2x+npi/2), from this, we can determine 50th dx/dy very fast and easy

 

[Galileo]

You could try focusing on the even derivatives:

0th: cos(2x)
2nd: -4cos(2x)
4th: 16cos(2x)
etc.

So you might guess for the (2n)th derivative:
$$(-1)^n4^ncos(2x)$$

and prove this with induction.