50th dy/dx

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. thats where im 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

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:
[tex](-1)^n4^ncos(2x)[/tex]

and prove this with induction.

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Alem2000	50th dy/dx Tweet 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. thats where im confused. My problem is with the negative signs. how can i catch that patteren.....

AKG Recognitions: Homework Help Science Advisor	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?

nta	50th dy/dx 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 Recognitions: Homework Help Science Advisor	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: [tex](-1)^n4^ncos(2x)[/tex] and prove this with induction.