Find the derivative of f(x) = x * cos(x + 3) + 7.

Step-by-step solution

1. 1

   Split the function into pieces. Differentiation is linear, so handle $x\cos(x+3)$ and the constant $7$ separately. The derivative of any constant is $0$, so the $+7$ contributes nothing — it shifts the graph vertically without changing any slope.

2. 2

   Set up the product rule on the first term. With $u = x$ and $v = \cos(x+3)$:

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

3. 3

   Differentiate each factor. $u' = 1$. For $v$, the chain rule gives

   $v' = -\sin(x+3)\cdot\frac{d}{dx}(x+3) = -\sin(x+3)\cdot 1 = -\sin(x+3)$

   The inner derivative is $1$ because the argument is a pure shift; had it been $\cos(3x+3)$ there would be an extra factor of $3$.

4. 4

   Assemble the derivative.

   $f'(x) = 1\cdot\cos(x+3) + x\left(-\sin(x+3)\right) = \cos(x+3) - x\sin(x+3)$

5. 5

   Check at a convenient point. At $x = -3$ the argument is $0$, so $f'(-3) = \cos 0 - (-3)\sin 0 = 1$. A central difference of $f$ at $x = -3$ with step $10^{-6}$ also returns $1.000000$. At $x = 0$: $f'(0) = \cos 3 = -0.989992$, matching the numerical derivative to six decimals.