Find the derivative of f(t) = t * sin(3t).

Step-by-step solution

1. 1

   Identify the structure as a product. $f$ is a product of $u = t$ and $v = \sin(3t)$, so the product rule applies:

   $\frac{d}{dt}\left[uv\right] = u'v + uv'$

   Neither factor can be simplified away first, so there is no shortcut around it.

2. 2

   Differentiate the simple factor.

   $u = t \quad\Longrightarrow\quad u' = 1$

3. 3

   Differentiate the composite factor with the chain rule. $\sin(3t)$ is the outer function $\sin$ applied to the inner function $3t$, whose derivative is $3$:

   $v = \sin(3t) \quad\Longrightarrow\quad v' = 3\cos(3t)$

   Forgetting that factor of $3$ is the single most common error here.

4. 4

   Assemble with the product rule.

   $f'(t) = (1)\sin(3t) + t\left(3\cos(3t)\right) = \sin(3t) + 3t\cos(3t)$

5. 5

   Verify numerically. At $t = 0.7$: the formula gives $\sin(2.1) + 2.1\cos(2.1) = 0.863209 - 1.060177 = -0.196967$. A central difference of $f$ at $t = 0.7$ with step $10^{-6}$ returns $-0.196967$ — agreement to six decimals. Note $f'(0) = \sin 0 + 0 = 0$, matching the fact that $t\sin 3t \approx 3t^2$ near the origin and so has a flat minimum there.