How was the chain rule used to find the derivative of a trigonometric function?

[PrudensOptimus]

I didn't understand how they arrived at this step:

lim as x -> 0 { 3 cos 3x (dy/dx) - 3 sin 3x (dy/dx)}

from

lim as x -> 0 { cos 3x (dy/dx) - sin3x (dy/dx) }

can some one explain pls?

 

[HallsofIvy]

I doubt that anyone can explain how one step follows from the other because you haven't shown the whole steps.

I assume what you have says that "lim as x -> 0 { 3 cos 3x (dy/dx) - 3 sin 3x (dy/dx)}" EQUALS something and that that follows from the fact that "lim as x -> 0 { cos 3x (dy/dx) - sin3x (dy/dx) }" equals something else. Unfortunately, you haven't told us what they equal.

Since the only difference between the two that I can see is that the first above is multiplied by 3, I would suspect that

"lim as x -> 0 { 3cos 3x (dy/dx) - 3sin3x (dy/dx) }= 3L"

follows from
"lim as x -> 0 { cos 3x (dy/dx) - sin3x (dy/dx) }= L" by multiplying the equation by 3. But of course, I can't be sure.

 

[PrudensOptimus]

they equal to -3 sin3x.

I was just wondering how did they get cos3x to be 3 cos 3x, sin 3x to be 3 sin 3x...

 

[gnome]

Please tell us exactly what it is that you (or "they") are trying to prove -- not just this one step, but the overall context.

And, what is the entire expression or equation you are starting out with?

 

[HallsofIvy]

No, they DON'T "equal to -3 sin3x." They can't because each has a
"lim as x->0". Once again, I can't say precisely what happened because you still haven't given us the whole thing but the obvious way to "get cos3x to be 3 cos 3x, sin 3x to be 3 sin 3x" is to multiply by 3!

 

[ecestud]

they equal to -3 sin3x.

I was just wondering how did they get cos3x to be 3 cos 3x, sin 3x to be 3 sin 3x...

To get that by getting the derivative of 3x which is 3 then copy cos3x and you will get
"3 cos 3x" same as "sin 3x to be 3 sin 3x". Not equal to -3 sin3x.

 

[stakehoagy]

if i understand what you mean, they just used the chain rule for derivatives.