Derivative of cos(3x) at x = 0: Is it 1?

[DB]

simple question, if the derivative of cos(x) as x->0 is 1, the is the derivative of cos(3x) as x->0 = 1 aswell?

 

[AKG]

The derivative of cos(x) as x->0 is -sin(x) as x->0 is -sin(0) is 0, not 1. But simply cos(x) as x->0 is cos(0) = 1. the derivative of cos(3x) as x->0 = -3sin(3x) as x->0 = -3sin(3*0) = -3sin(0) = -3*0 = 0. cos(3x) as x->0 = cos(3*0) = cos(0) = 1. The derivative of cos(x) as x->a = -sin(x) as x->a = -sin(a). The derivative of cos(3x) as x->a = -3sin(3x) as x->a = -3sin(3a).

 

[mathman]

The derivative of cos(x) is -sin(x).

 

[AKG]

Yeah, I knew that. Edited.

 

[moose]

Think about what a derivative means. This should be plenty enough to give you the answer you're looking for.

 

[HallsofIvy]

simple question, if the derivative of cos(x) as x->0 is 1, the is the derivative of cos(3x) as x->0 = 1 aswell?

The derivative of sin x goes to 1 as x goes to 0. No, the derivative of sin(3x) does not go to 1. It goes to 3. Do you see why?

 

[cDimino]

Curious... it looks like you're looking for a limit. I wasn't aware that derivitaves "approached" anything.

 

[HallsofIvy]

Curious... it looks like you're looking for a limit. I wasn't aware that derivitaves "approached" anything.

The derivative of a function of x is a function of x and can approach a limit as well as any function.

 

[moose]

The derivative tells you how the result will change with a change in the x. If it were to linearly change, the derivative would say how y changes when 1 is added to x. So, if you have cos(3x), the derivative will have to be 3 times greater than cosx, because if you raise x by an amount (say, 1), you are raising the quantity inside the cos by 3 times that amount. It's the same concept as the derivative (slope) of y=x and y=3x.

This is how the chain rule can be derived...

 

[Robokapp]

just U-substitute u=3x if you must and apply chain rule. dy/dx=dy/du*du/dx