If `y=cos2x*sin3x," then "(d^(2)y)/(dx^(2))=`

Options

A. `-(1)/(2)\(5cos5x+sinx)`
B. `-(1)/(2)\(25sin5x-sinx)`
C. `(1)/(2)\(25cos5x-sinx)`
D. `-(1)/(2)\(25sin5x+sinx)`

Answer

To find the second derivative of the function \( y = \cos(2x) \sin(3x) \), we can follow these steps:

### Step 1: Rewrite the function using the product-to-sum identities
We can use the product-to-sum identity for \( 2 \cos A \sin B \):
\[ 2 \cos A \sin B = \sin(A + B) - \sin(A - B) \]
In our case, \( A = 2x \) and \( B = 3x \). Thus, we can write:
\[ y = \cos(2x) \sin(3x) = \frac{1}{2} \left( \sin(5x) - \sin(-x) \right) \]
Since \( \sin(-x) = -\sin(x) \), we have:
\[ y = \frac{1}{2} \left( \sin(5x) + \sin(x) \right) \]

### Step 2: Find the first derivative \( \frac{dy}{dx} \)
Now, we differentiate \( y \):
\[ \frac{dy}{dx} = \frac{1}{2} \left( \frac{d}{dx}[\sin(5x)] + \frac{d}{dx}[\sin(x)] \right) \]
Using the chain rule, we find:
\[ \frac{d}{dx}[\sin(5x)] = 5 \cos(5x) \quad \text{and} \quad \frac{d}{dx}[\sin(x)] = \cos(x) \]
Thus,
\[ \frac{dy}{dx} = \frac{1}{2} \left( 5 \cos(5x) + \cos(x) \right) \]

### Step 3: Find the second derivative \( \frac{d^2y}{dx^2} \)
Now we differentiate \( \frac{dy}{dx} \):
\[ \frac{d^2y}{dx^2} = \frac{1}{2} \left( \frac{d}{dx}[5 \cos(5x)] + \frac{d}{dx}[\cos(x)] \right) \]
Using the chain rule again:
\[ \frac{d}{dx}[5 \cos(5x)] = -25 \sin(5x) \quad \text{and} \quad \frac{d}{dx}[\cos(x)] = -\sin(x) \]
Thus,
\[ \frac{d^2y}{dx^2} = \frac{1}{2} \left( -25 \sin(5x) - \sin(x) \right) \]
This simplifies to:
\[ \frac{d^2y}{dx^2} = -\frac{1}{2} (25 \sin(5x) + \sin(x)) \]

### Final Answer
\[ \frac{d^2y}{dx^2} = -\frac{1}{2} (25 \sin(5x) + \sin(x)) \]