If `y= e^(4x) cos 5x ,then (d^(2)y)/(dx^(2))=`

Options

A. ` -e^(4x) (9cos 5x +20 sin 5x )`
B. ` e^(4x) (9cos 5x+ 20sin 5x)`
C. `- e^(4x) (9cos 5x+ 40sin 5x)`
D. ` e^(4x) (9cos 5x+ 40sin 5x)`

Answer

To find the second derivative of the function \( y = e^{4x} \cos(5x) \), we will follow these steps:

### Step 1: Find the first derivative \( \frac{dy}{dx} \)

We will use the product rule for differentiation, which states that if \( y = u \cdot v \), then \( \frac{dy}{dx} = u'v + uv' \).

Let:
- \( u = e^{4x} \) and \( v = \cos(5x) \)

Now, we differentiate \( u \) and \( v \):
- \( u' = \frac{d}{dx}(e^{4x}) = 4e^{4x} \)
- \( v' = \frac{d}{dx}(\cos(5x)) = -5\sin(5x) \)

Now applying the product rule:
\[ \frac{dy}{dx} = u'v + uv' = (4e^{4x})(\cos(5x)) + (e^{4x})(-5\sin(5x)) \]
\[ \frac{dy}{dx} = 4e^{4x} \cos(5x) - 5e^{4x} \sin(5x) \]

### Step 2: Find the second derivative \( \frac{d^2y}{dx^2} \)

Now we need to differentiate \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = 4e^{4x} \cos(5x) - 5e^{4x} \sin(5x) \]

Using the product rule again on both terms:

1. Differentiate \( 4e^{4x} \cos(5x) \):
- Let \( u_1 = 4e^{4x} \) and \( v_1 = \cos(5x) \)
- \( u_1' = 16e^{4x} \)
- \( v_1' = -5\sin(5x) \)
\[ \frac{d}{dx}(4e^{4x} \cos(5x)) = u_1'v_1 + u_1v_1' = (16e^{4x})(\cos(5x)) + (4e^{4x})(-5\sin(5x)) \]
\[ = 16e^{4x} \cos(5x) - 20e^{4x} \sin(5x) \]

2. Differentiate \( -5e^{4x} \sin(5x) \):
- Let \( u_2 = -5e^{4x} \) and \( v_2 = \sin(5x) \)
- \( u_2' = -20e^{4x} \)
- \( v_2' = 5\cos(5x) \)
\[ \frac{d}{dx}(-5e^{4x} \sin(5x)) = u_2'v_2 + u_2v_2' = (-20e^{4x})(\sin(5x)) + (-5e^{4x})(5\cos(5x)) \]
\[ = -20e^{4x} \sin(5x) - 25e^{4x} \cos(5x) \]

### Step 3: Combine the results

Now we combine the results from both differentiations:
\[ \frac{d^2y}{dx^2} = (16e^{4x} \cos(5x) - 20e^{4x} \sin(5x)) + (-20e^{4x} \sin(5x) - 25e^{4x} \cos(5x)) \]

Combining like terms:
\[ = (16e^{4x} - 25e^{4x}) \cos(5x) + (-20e^{4x} - 20e^{4x}) \sin(5x) \]
\[ = -9e^{4x} \cos(5x) - 40e^{4x} \sin(5x) \]

### Final Answer
Thus, the second derivative is:
\[ \frac{d^2y}{dx^2} = -e^{4x}(9 \cos(5x) + 40 \sin(5x)) \]