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

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

First of all, let's recall the , 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)