Thread: Derivative proof using sine and cosine

Hi guys can you please help me with this problem:

If n is a positive integer, prove this derivative:

(sin^n)xcosnx = n(sin^n-1)xcos(n+1)x

how the heck do I do that?

Originally Posted by RezMan

Hi guys can you please help me with this problem:

If n is a positive integer, prove this derivative:

(sin^n)xcosnx = n(sin^n-1)xcos(n+1)x

how the heck do I do that?

First, is your problem


where


represents the nth derivative of sin(x)? If so, note that the theorem is not true for n = 1.

-Dan

My interpretation, though it is written badly, is to show that the derivative of f(x)= sin^n(x)cos(nx) is f'(x)= n sin^{n-1}(x)cos((n+1)x).

To start with use the product rule and chain rule:
f'(x)= n sin^{n-1}(x)cos(x)cos(nx)+ sin^n(x)(-n sin(nx))= n sin^{n-1}(x)[cos(x)cos(nx)- sin(x)sin(nx)]

Now use the trig identity cos(a+b)= cos(a)cos(b)- sin(a)sin(b) to rewrite the quantitiy inside the [ ].

As psychic as ever. Good call.

-Dan

Thank you so much! But I still need some help. Please bear with me. Also, topsquark, can you show me how you made the equation look like that on this board?

I get that (sin^n)x turns into n(sin^n-1) but can you explain in more detail what's going on here cos(x)cos(nx)+ sin^n(x)(-n sin(nx)). Like I don't quite see where the chain rule is applied.

Originally Posted by RezMan

Also, topsquark, can you show me how you made the equation look like that on this board?

You can visit the Latex Help forum to learn how to use the LaTeX commands. For the moment LaTeX is down but there is a temporary fix that I like. Instead of using the [tex] tags, use [tex].

-Dan

thanks guys!