If `y=e^(-x)(6 cos ax+5 sin bx)` then `dy/dx`=

Options

A. `dy/dx=e^(-x)[-6asinax+5bcosbx-6cosax-5sinbx]`
B. `dy/dx=e^(-x)[6asinax-5bcosbx-6cosax-5sinbx]`
C. `dy/dx=e^(-x)[-6asinax+5bcosbx+6cosax+5sinbx]`
D. None of these

Answer

To find the derivative of the function \( y = e^{-x}(6 \cos(ax) + 5 \sin(bx)) \), we will use the product rule of differentiation. The product rule states that if \( y = u \cdot v \), then the derivative \( \frac{dy}{dx} = u'v + uv' \), where \( u' \) and \( v' \) are the derivatives of \( u \) and \( v \) respectively.

### Step 1: Identify \( u \) and \( v \)
Let:
- \( u = e^{-x} \)
- \( v = 6 \cos(ax) + 5 \sin(bx) \)

### Step 2: Differentiate \( u \) and \( v \)
1. Differentiate \( u \):
\[ u' = \frac{d}{dx}(e^{-x}) = -e^{-x} \]

2. Differentiate \( v \):
\[ v' = \frac{d}{dx}(6 \cos(ax) + 5 \sin(bx)) = 6 \cdot (-\sin(ax) \cdot a) + 5 \cdot \cos(bx) \cdot b \]
Simplifying this gives:
\[ v' = -6a \sin(ax) + 5b \cos(bx) \]

### Step 3: Apply the product rule
Now, apply the product rule:
\[ \frac{dy}{dx} = u'v + uv' \]
Substituting the values we found:
\[ \frac{dy}{dx} = (-e^{-x})(6 \cos(ax) + 5 \sin(bx)) + (e^{-x})(-6a \sin(ax) + 5b \cos(bx)) \]

### Step 4: Factor out \( e^{-x} \)
We can factor out \( e^{-x} \):
\[ \frac{dy}{dx} = e^{-x} \left( - (6 \cos(ax) + 5 \sin(bx)) - 6a \sin(ax) + 5b \cos(bx) \right) \]

### Step 5: Combine the terms
Combining the terms inside the parentheses:
\[ \frac{dy}{dx} = e^{-x} \left( -6 \cos(ax) - 5 \sin(bx) - 6a \sin(ax) + 5b \cos(bx) \right) \]

### Final Answer
Thus, the derivative \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = e^{-x} \left( -6 \cos(ax) - 5 \sin(bx) - 6a \sin(ax) + 5b \cos(bx) \right) \]