Find the n -th derivative of y = cos x

Find the $n$-th derivative of $y = \cos x$.

Step 1: The first four derivatives

\[ y' = -\sin x, \qquad y'' = -\cos x, \qquad y''' = \sin x, \qquad y^{\mathrm{IV}} = \cos x. \]

The fourth derivative returns to the start, so the sequence is periodic with period four.

Step 2: Rewrite each as a shifted cosine

The identity $-\sin x = \cos\!\left(x+\tfrac{\pi}{2}\right)$ suggests reading every step as a shift: \[y' = \cos\!\left(x+\frac{\pi}{2}\right), \quad y'' = \cos\bigl(x+\pi\bigr), \quad y''' = \cos\!\left(x+\frac{3\pi}{2}\right), \quad y^{\mathrm{IV}} = \cos\bigl(x+2\pi\bigr).\]

Step 3: The general formula, by induction

Claim: $y^{(k)} = \cos\!\left(x+\dfrac{k\pi}{2}\right)$. The base case is the first line of Step 2. Assuming it for $k$: \[y^{(k+1)} = -\sin\!\left(x+\frac{k\pi}{2}\right) = \cos\!\left(x+\frac{k\pi}{2}+\frac{\pi}{2}\right) = \cos\!\left(x+\frac{\bigl(k+1\bigr)\pi}{2}\right).\]

Step 4: Verification

\[\begin{array}{c|c|c} n\bmod4 & \text{formula} & \text{direct} \\ \hline 0 & \cos x & \cos x \\ 1 & \cos\!\left(x+\tfrac{\pi}{2}\right) = -\sin x & -\sin x \\ 2 & \cos\bigl(x+\pi\bigr) = -\cos x & -\cos x \\ 3 & \cos\!\left(x+\tfrac{3\pi}{2}\right) = \sin x & \sin x \end{array}\]

\[\text{All four rows agree} \;\;\checkmark\]

\[y^{(n)}(0) = \cos\frac{n\pi}{2} = 1,0,-1,0,\ldots, \text{ matching } \cos x = 1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}-\cdots \;\;\checkmark\]