How Do You Write the nth Derivative Formula for f(x) = sin(3x)?

[Lizabeth54]

How would the equation be written for the nth derivative? I understand it will be different for odd dertivatives and even derivatives. I'm working on f(x)= sin (3x) and I cannot figure out how to write the equation to the nth derivative.
f'(x) = 3cos(3x) 1st derivative
f"(x) = -9sin(3x) 2nd derivative
f"'(x) = -27cos(3x) 3rd derivative
f""(x) = 81sin(3x) 4th derivative.
so for the odd nth derivatives of f(x), I have come up with something like:
(-1) 3^n cos(3x)

and for the even nth derivatives of f(x), I have come up with something like:
(-1) 3^n sin(3x)

but it's just not right. How do I get alternating signs in this formula?

 

[d_leet]

How would the equation be written for the nth derivative? I understand it will be different for odd dertivatives and even derivatives. I'm working on f(x)= sin (3x) and I cannot figure out how to write the equation to the nth derivative.
f'(x) = 3cos(3x) 1st derivative
f"(x) = -9sin(3x) 2nd derivative
f"'(x) = -27cos(3x) 3rd derivative
f""(x) = 81sin(3x) 4th derivative.
so for the odd nth derivatives of f(x), I have come up with something like:
(-1) 3^n cos(3x)

and for the even nth derivatives of f(x), I have come up with something like:
(-1) 3^n sin(3x)

but it's just not right. How do I get alternating signs in this formula?

Well you've already dealt with the coefficient and the fact that it alternates between sine and cosine on even and odd derivatives respectively, so maybe you have noticed that it cycles back to the same function ignoring the coefficient every fourth derivative, so think about there being 4 cases depending on what derivative you're taking.. do you see what I'm getting at?

 

[Lizabeth54]

Yeah, I understand what you mean about it repeating every 4. I am looking for just 2 functions... and I was given a hint that it would have to do with the exponent of 3 but whatever I try, nothing seems to work out perfect universally... I think there's something to do with the "3^n" that I can manipulate to make this automatically have alternating signs at any number n...?

 

[d_leet]

Yeah, I understand what you mean about it repeating every 4. I am looking for just 2 functions... and I was given a hint that it would have to do with the exponent of 3 but whatever I try, nothing seems to work out perfect universally... I think there's something to do with the "3^n" that I can manipulate to make this automatically have alternating signs at any number n...?

Ohhh I see it now, ok well if you raise 3 to an even exponent you can express it as 3^2 or 9 to an exponent right, so you can then make 2 cases fairly easily for even and odd derivatives that are easier to make alternatingly positive or negative.. It might take a bit of thought to figure out how to get alternating signs for the odd cases but it is possible.

 

[HallsofIvy]

Another way is to note that every even number can be written "2n" and every odd number "2n+1". Take a look at the signs of the derivative in terms of that n.