Find y' for y = sin 3x
Find $y'$ for $y = \sin 3x$.
Explanation of Method
The simplest chain rule: a linear inner function.
Step 1: Apply the chain rule
With outer function $\sin u$ and inner function $u = 3x$, so that $u' = 3$: \[y' = \cos u\cdot u' = \cos(3x)\cdot 3 = 3\cos 3x.\]
Step 2: Verification
\[x = 0:\quad y'(0) = 3\cos 0 = 3; \qquad \frac{\sin(3\cdot 10^{-5})}{10^{-5}} \approx 3.0 \;\;\checkmark\]
\[y' = 0 \;\Longleftrightarrow\; 3x = \frac{\pi}{2}+k\pi \;\Longleftrightarrow\; x = \frac{\pi}{6}+\frac{k\pi}{3} \;\;\checkmark\]
Compressing the sine horizontally by a factor of $3$ multiplies every slope by $3$, which is exactly what the factor $u' = 3$ records.
Result
Final answer:
\[\boxed{\,y' = 3\cos 3x\,}\]
Remark
Remark. The common mistake is to write $(\sin 3x)' = \cos 3x$, forgetting the inner derivative. A quick sanity check catches it: $\sin 3x$ completes three full oscillations while $\sin x$ completes one, so its slopes must be three times as steep.