I'm not too familiar with doing the derivative of trig functions
say you had
sin(4x)^2
Would that be?
8sin(4x)cos(4x)

I'm not too familiar with doing the derivative of trig functions
say you had
sin(4x)^2
Would that be?
8sin(4x)cos(4x)
You could mean sin<sup>2</sup> (4x) or sin ( (4x)<sup>2</sup> ) which would of course be different derivitives.
Lets first deal with sin<sup>2</sup> (4x). If you think of it like a function within a function it may be easier. First lets start with the more simple:
(d/dx)( sin x ) = cos x
Here are some examples of exponents of sin and their derivitives. These are done using the chain rule. The derivitive of x<sup>3</sup> is 3x<sup>2</sup> and the derivitive of sin x is cos x, which is the inside function.
(d/dx)( sin x)<sup>2</sup> = 2 sin<sup>1</sup> x cos x
(d/dx)( sin x)<sup>3</sup> = 3 sin<sup>2</sup> x cos x
(d/dx)( sin x)<sup>4</sup> = 4 sin<sup>3</sup> x cos x
(d/dx)( sin x)<sup>5</sup> = 5 sin<sup>4</sup> x cos x
We could apply the same concept to sin 4x as well, but since there is no exponent, our sine element will disapear leaving only the cosine part like so:
(d/dx)( sin 4x) = 4 cos 4x
(d/dx)( sin 5x) = 5 cos 5x
(d/dx)( sin 6x) = 6 cos 6x
What we are doing here is first taking the derivitive of the outer function sin (f(z)) lets say, and then mulitiplying our answer by the derivitive of f(z).
What I used to do when I was having problems figuring out derivitives of "multiple chains" as I might think of them, was to write out their equivilant in terms of functions. I think I have a lot of examples of this type of work in my Chapter 3 thread.
I'm not sure that my descriptions here were very good. If you have more questions, just ask away
Oh, and the answer to your question, unless I am mistaken, is yes, that is correct.
yep perfectly correct
Thank you for the help
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