# The nth derivative.

1. Jul 16, 2006

### vin-math

The nth derivative........

I dont know how to proof the nth derivative of cos ax with respect to a is a^ncos (ax+ na/2). Can any one help me?

2. Jul 16, 2006

### arildno

Well, it isn't. Where's your pi?

3. Jul 16, 2006

### vin-math

ooops, i type wrongly. it should be "proof the nth derivative of cos (pi)x with respect to a is (pi)^ncos [(pi)x+ n(pi)/2)]"
actually i dont know hot to type the notation of pi here, so i replace the pi with a

4. Jul 16, 2006

### d_leet

I think you could prove this by induction, show it's true for n=1, assume it's true for n=k then show that it being true for n=k implies it being true for n=k+1.

5. Jul 16, 2006

### vin-math

But can i just calculate the final answer (the nth derivative), apart from using the mathematical induction?

6. Jul 16, 2006

### HallsofIvy

Staff Emeritus
You might want to keep in mind that
$$cos(x+ \frac{n\pi}{2})$$
is equal to
$$(-1)^{\frac{n}{2}}cos(x)$$
if n is even,
$$(-1)^{\frac{n-1}{2}}sin(x)$$
if n is odd.

7. Jul 16, 2006

### d_leet

What do you mean? The way you stated the problem was that you didn't know how to prove it, and it seems that the best way to prove this would be to use mathematical induction and HallsofIvy's suggestion.

8. Jul 16, 2006

### vin-math

thx!
I think i now hv some idea in solve the Q.

9. Jul 16, 2006

### vin-math

I mean that like proofing identities, we will proof the L.H.S equal to the R.H.S. I want to do the same thing in this Q.

10. Jul 16, 2006

### d_leet

And how do you expect to do this? I don't really think that you can do something like that in this situation, and so I will again recommend that you try to prove this using mathematical induction, it really doesn't seem like it would be that hard or long of a proof given the suggestion that HallsofIvy gave you.

11. Jul 17, 2006

### lurflurf

I pressume the result you intend is something like
(0)
$$( \frac{\d}{\d x} )^n \cos(a x+b)=a^n \cos(a x+b+n \frac{\pi}{2})$$
use the following basic results
(1) An obvious trigonometry identity
$$\frac{\Delta}{\Delta x} \cos(a x+b)=a \cos(a x+b+a \frac{h}{2}+\frac{\pi}{2})\frac{\sin(h/2)}{h/2}$$
(2) Continuity of cosine
$$\lim_{h->0}(\cos(x+h)-\cos(x))=0$$
(3) sin(x)~x x small
$$\lim_{h->0}\frac{\sin(h)}{h}=1$$
(4) apply (1) n times
$$(\frac{\Delta}{\Delta x})^n \cos(a x+b)=a \cos(a x+b+a n \frac{h}{2}+n \frac{\pi}{2})(\frac{\sin(h/2)}{h/2})^n$$
(5) passage to the limit (h->0) of (4) using (2) and (3) give the desired result
$$( \frac{\d}{\d x} )^n \cos(a x+b)=a^n \cos(a x+b+n \frac{\pi}{2})$$

Last edited: Jul 17, 2006
12. Jul 17, 2006

### HallsofIvy

Staff Emeritus
By the way, you keep saying "the derivative with respect to a". If that's what you really mean, then the answer you give is NOT correct. But everyone has been assuming that you meant "with respect to x" anyway.

13. Jul 18, 2006

### vin-math

Yes, u are right

14. Jul 18, 2006

### lurflurf

I pressume the result you intend is something like
(0)
$$( \frac{d}{dx} )^n \cos(a x+b)=a^n \cos(a x+b+n \frac{\pi}{2})$$
use the following basic results
(1) An obvious trigonometry identity
$$\frac{\Delta}{\Delta x} \cos(a x+b)=a \cos(a x+b+a \frac{h}{2}+\frac{\pi}{2})\frac{\sin(a h/2)}{a h/2}$$
(2) Continuity of cosine
$$\lim_{h->0}(\cos(x+h)-\cos(x))=0$$
(3) sin(x)~x x small
$$\lim_{h->0}\frac{\sin(h)}{h}=1$$
(4) apply (1) n times
$$(\frac{\Delta}{\Delta x} )^n \cos(a x+b)=a^n \cos(a x+b+a n \frac{h}{2}+n \frac{\pi}{2})(\frac{\sin(a h/2)}{a h/2})^n$$
(5) passage to the limit (h->0) of (4) using (2) and (3) give the desired result
$$( \frac{d}{dx} )^n \cos(a x+b)=a^n \cos(a x+b+n \frac{\pi}{2})$$