Find the first and second derivatives of f(x) = e^(-x^2) * sin(3x).

Step-by-step solution

1. 1

   Set up the product rule. With $u = e^{-x^2}$ and $v = \sin(3x)$, both factors are themselves composites, so the chain rule is needed inside each derivative:

   $u' = e^{-x^2}\cdot(-2x) = -2xe^{-x^2}, \qquad v' = 3\cos(3x)$

   The inner derivatives $-2x$ and $3$ are the two factors most often dropped here.

2. 2

   Assemble the first derivative and factor.

   $f'(x) = -2xe^{-x^2}\sin(3x) + 3e^{-x^2}\cos(3x) = e^{-x^2}\left(3\cos 3x - 2x\sin 3x\right)$

   Pulling $e^{-x^2}$ out front keeps the second differentiation manageable.

3. 3

   Differentiate the bracket for the second derivative. Let $w = 3\cos 3x - 2x\sin 3x$. Then

   $w' = -9\sin 3x - \left(2\sin 3x + 6x\cos 3x\right) = -11\sin 3x - 6x\cos 3x$

   The $-2x\sin 3x$ term needs the product rule again, contributing both $-2\sin 3x$ and $-6x\cos 3x$.

4. 4

   Apply the product rule once more.

   $f''(x) = u'w + uw' = -2xe^{-x^2}\left(3\cos 3x - 2x\sin 3x\right) + e^{-x^2}\left(-11\sin 3x - 6x\cos 3x\right)$

5. 5

   Collect the sine and cosine terms. The cosine terms give $-6x - 6x = -12x$ and the sine terms give $4x^2 - 11$:

   $f''(x) = e^{-x^2}\left(\left(4x^2 - 11\right)\sin 3x - 12x\cos 3x\right)$

6. 6

   Check at x = 0. From the formulas, $f'(0) = 1\cdot(3 - 0) = 3$ and $f''(0) = 1\cdot(-11\cdot 0 - 0) = 0$. Directly, near $0$ we have $f(x) \approx (1-x^2)(3x - \tfrac{9x^3}{2}) = 3x + O(x^3)$, so the slope at the origin is $3$ and the second derivative vanishes — both confirmed.