Find the second and third derivatives of y = sin x
Find the second and third derivatives of $y = \sin x$.
Explanation of Method
The first two steps of a cycle that repeats every four.
Step 1: Differentiate three times in turn
\[ y' = \cos x, \qquad y'' = \bigl(\cos x\bigr)' = -\sin x, \qquad y''' = \bigl(-\sin x\bigr)' = -\cos x. \]
Step 2: The pattern
A fourth differentiation returns to the start, so the derivatives run in a cycle of length four: \[\sin x \;\longrightarrow\; \cos x \;\longrightarrow\; -\sin x \;\longrightarrow\; -\cos x \;\longrightarrow\; \sin x \;\longrightarrow\; \cdots\]
Each step is a shift of the argument by $\tfrac{\pi}{2}$, which gives the compact formula
\[ y^{(n)} = \sin\!\left(x+\frac{n\pi}{2}\right). \]
Step 3: Verification
\[n = 2:\ \sin\bigl(x+\pi\bigr) = -\sin x \;\;\checkmark\]
\[n = 3:\ \sin\!\left(x+\frac{3\pi}{2}\right) = -\cos x \;\;\checkmark\]
\[n = 4:\ \sin\bigl(x+2\pi\bigr) = \sin x \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y'' = -\sin x, \qquad y''' = -\cos x\,}\]
Remark
Remark. The relation $y'' = -y$ is the defining property of the sine: it is the equation of simple harmonic motion, and $\sin x$ and $\cos x$ are its two independent solutions. Every fact about repeated differentiation of trigonometric functions ultimately follows from that one identity.