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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?
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 a function of x is a function of x and can approach a limit as well as any function.cDimino said:Curious... it looks like you're looking for a limit. I wasn't aware that derivitaves "approached" anything.
The derivative of cos(3x) at x = 0 is -3sin(3x).
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).
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).
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).
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).