The problem asks to differentiate $y = 2\sin x + 3\cos x$.

Step-by-step solution:

1. The derivative of $\sin x$ with respect to $x$ is $\cos x$.
2. The derivative of $\cos x$ with respect to $x$ is $-\sin x$.

So, applying these derivatives:

[ \frac{dy}{dx} = \frac{d}{dx}[2\sin x] + \frac{d}{dx}[3\cos x] ]

Calculation:

$\frac{dy}{dx} = 2\cos x - 3\sin x$

Final Answer:

$\boxed{\frac{dy}{dx} = 2\cos x - 3\sin x}$

Would you like further details or examples? Here are related questions to expand understanding:

1. How do trigonometric derivatives apply in real-world problems?
2. Can you integrate $2\cos x - 3\sin x$ to verify the result?
3. What is the derivative of $y = 2\sin(2x) + 3\cos(2x)$?
4. How would the derivative change if $y$ included a $\tan x$ term?
5. Can this differentiation process apply to complex functions involving $\sin x$ and $\cos x$?

Tip: Memorize the derivatives of basic trigonometric functions—they are