- #1

PrudensOptimus

- 641

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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?

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- Thread starter PrudensOptimus
- Start date

- #1

PrudensOptimus

- 641

- 0

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?

- #2

HallsofIvy

Science Advisor

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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.

- #3

PrudensOptimus

- 641

- 0

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

- #4

gnome

- 1,037

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And, what is the

- #5

HallsofIvy

Science Advisor

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"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!

- #6

ecestud

- 1

- 0

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.

Last edited:

- #7

stakehoagy

- 29

- 0

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

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