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

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In summary, the derivative of cos(x) is -sin(x) and the derivative of cos(3x) is -3sin(3x). The derivative of a function of x can approach a limit as well, and in this case, the derivative of cos(3x) as x->0 will approach 0, not 1. The chain rule can be applied to find the derivative of cos(3x) by substituting u=3x and applying the chain rule.
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DB
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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?
 
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  • #2
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).
 
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  • #3
The derivative of cos(x) is -sin(x).
 
  • #4
Yeah, I knew that. :redface: Edited.
 
  • #5
Think about what a derivative means. This should be plenty enough to give you the answer you're looking for.
 
  • #6
DB said:
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?
 
  • #7
Curious... it looks like you're looking for a limit. I wasn't aware that derivitaves "approached" anything.
 
  • #8
cDimino said:
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.
 
  • #9
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...
 
  • #10
just U-substitute u=3x if you must and apply chain rule. dy/dx=dy/du*du/dx
 

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

What is the derivative of cos(3x) at x = 0?

The derivative of cos(3x) at x = 0 is -3sin(3x).

How do you calculate the derivative of cos(3x) at x = 0?

The derivative of cos(3x) at x = 0 can be calculated using the chain rule, which states that the derivative of f(g(x)) is equal to f'(g(x)) * g'(x). In this case, f(x) = cos(x) and g(x) = 3x. Therefore, the derivative is equal to -3sin(3x).

Why is the derivative of cos(3x) at x = 0 equal to -3sin(3x)?

The derivative of cos(3x) is equal to -3sin(3x) because the derivative of cos(x) is equal to -sin(x), and the derivative of 3x is equal to 3. When these two functions are combined using the chain rule, the result is -3sin(3x).

Can you prove that the derivative of cos(3x) at x = 0 is -3sin(3x)?

Yes, the derivative of cos(3x) at x = 0 can be proven using the definition of the derivative, which states that the derivative of a function f(x) at a point x = a is equal to the limit as h approaches 0 of [f(a + h) - f(a)] / h. By plugging in x = 0 and evaluating the limit, the result is -3sin(3x).

Is the derivative of cos(3x) at x = 0 always equal to -3sin(3x)?

Yes, the derivative of cos(3x) at x = 0 is always equal to -3sin(3x). This is because the derivative of cosine is always equal to negative sine, and the coefficient of x (in this case, 3) remains the same when taking the derivative. Therefore, the derivative will always be -3sin(3x).

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