1. **Problem:** Find the $n$th derivative of the function $f(x) = \cos(ax + b)$.\n\n2. **Formula and rules:** The derivative of $\cos(u)$ with respect to $x$ is $-\sin(u) \cdot \frac{du}{dx}$. For higher order derivatives, we use the fact that derivatives of sine and cosine functions cycle every four derivatives: \n$$\frac{d}{dx}\cos(u) = -\sin(u)u'$$\n$$\frac{d^2}{dx^2}\cos(u) = -\cos(u)(u')^2 - \sin(u)u''$$\nBut since $u = ax + b$ is linear, $u' = a$ and $u'' = 0$. So the derivatives simplify.\n\n3. **Step-by-step:**\n- First derivative: $$f'(x) = -a \sin(ax + b)$$\n- Second derivative: $$f''(x) = -a^2 \cos(ax + b)$$\n- Third derivative: $$f^{(3)}(x) = a^3 \sin(ax + b)$$\n- Fourth derivative: $$f^{(4)}(x) = a^4 \cos(ax + b)$$\n\n4. **Pattern:** The derivatives repeat every 4 steps with alternating signs and sine/cosine functions. The $n$th derivative is: $$f^{(n)}(x) = a^n \cos\left(ax + b + n\frac{\pi}{2}\right)$$\nThis uses the phase shift property of cosine and sine.\n\n5. **Final answer:**\n$$f^{(n)}(x) = a^n \cos\left(ax + b + n\frac{\pi}{2}\right)$$\nThis formula gives the $n$th derivative of $f(x) = \cos(ax + b)$ for any integer $n \geq 0$.