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?

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- May 1st 2011, 04:09 AMRezManDerivative 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? - May 1st 2011, 04:27 AMtopsquark
- May 1st 2011, 05:04 AMHallsofIvy
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 [ ]. - May 1st 2011, 05:11 AMtopsquark
As psychic as ever. (Nod) Good call.

-Dan - May 1st 2011, 09:32 PMRezMan
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. - May 2nd 2011, 02:07 AMtopsquark
- May 5th 2011, 07:28 PMRezMan
thanks guys!