Question

What is the formula for the nth derivative of `sin (ax +b)` and `cos(ax+b)`?

Answer 1

Let's make some explicit computations, and from there let's try to find a general formula.

First of all, let's recall the chain rule, which states that if you have a composite function `f(g(x))`, then `(f(g(x))'=f'(g(x))*g'(x)`

In your case, `f(x)=\sin(x)` (or `\cos(x)`), and `g(x)=ax+b`. So, we can easily observe that `g'(x)=a`. We have thus a first, important result: every time we derive, we will have a factor `a` to multiply our expression. As for the trigonometric part, we know that trigonometric functions have this "derivative loop":
`\sin(x)\rightarrow \cos(x)\rightarrow -\sin(x)\rightarrow -\cos(x)\rightarrow \sin(x)...`

We are ready for the final answer: every time we derive, we know which trigonometric function will appear, and we also know that there will be a certain power `a^n` to multiply, where `n` is the number of derivatives taken so far. In formulas (we are assuming of course that `n\geq 0`, where `n=0` means doing no derivatives):

`{d^n}/{dx} \sin(ax+b) = a^n F_n(ax+b)`, where `F_n(x)` is a function which equals:

1. `\sin(x)` if `n=4k` (i.e., `n` is a multiple of 4 (including `n=0`))
2. `\cos(x)` if `n=4k+1` (i.e., `n` is one unit away from a multiple of 4)
3. `-\sin(x)` if `n=4k+2` (i.e., `n` is two units away from a multiple of 4)
4. `-\cos(x)` if `n=4k+3` (i.e., `n` is three units away from a multiple of 4)

As for the cosine function, the logic is identical, you only need to shift the values of the function I called `F_n(x)`, because you start from the second point of that "derivative loop". So, you will have

`{d^n}/{dx} \cos(ax+b) = a^n F_n(ax+b)`, where `F_n(x)` is a function which equals:

1. `\cos(x)` if `n=4k` (i.e., `n` is a multiple of 4 (including `n=0`))
2. `-\sin(x)` if `n=4k+1` (i.e., `n` is one unit away from a multiple of 4)
3. `-\cos(x)` if `n=4k+2` (i.e., `n` is two units away from a multiple of 4)
4. `\sin(x)` if `n=4k+3` (i.e., `n` is three units away from a multiple of 4)