Find the second derivative of:
`(i)x sin x" "(ii)e^(2x)cos 3x" "(iii)x^(3)logx`

Answer

To find the second derivative of the given functions, we will follow the steps of differentiation carefully for each part.

### Part (i): \( y = x \sin x \)

**Step 1: First Derivative**
Using the product rule:
\[ \frac{dy}{dx} = u'v + uv' \]
where \( u = x \) and \( v = \sin x \).

Calculating \( u' \) and \( v' \):
- \( u' = 1 \)
- \( v' = \cos x \)

Now applying the product rule:
\[ \frac{dy}{dx} = (1)(\sin x) + (x)(\cos x) = \sin x + x \cos x \]

**Step 2: Second Derivative**
Now we differentiate \( \frac{dy}{dx} \):
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}(\sin x + x \cos x) \]
Using the product rule again for \( x \cos x \):
\[ \frac{d^2y}{dx^2} = \cos x + \left(1 \cdot \cos x + x \cdot (-\sin x)\right) \]
Combining terms:
\[ \frac{d^2y}{dx^2} = \cos x + \cos x - x \sin x = 2\cos x - x \sin x \]

### Part (ii): \( y = e^{2x} \cos(3x) \)

**Step 1: First Derivative**
Using the product rule:
\[ \frac{dy}{dx} = u'v + uv' \]
where \( u = e^{2x} \) and \( v = \cos(3x) \).

Calculating \( u' \) and \( v' \):
- \( u' = 2e^{2x} \)
- \( v' = -3\sin(3x) \)

Now applying the product rule:
\[ \frac{dy}{dx} = (2e^{2x})(\cos(3x)) + (e^{2x})(-3\sin(3x)) = 2e^{2x} \cos(3x) - 3e^{2x} \sin(3x) \]

**Step 2: Second Derivative**
Now we differentiate \( \frac{dy}{dx} \):
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}(2e^{2x} \cos(3x) - 3e^{2x} \sin(3x)) \]
Using the product rule on both terms:
\[ \frac{d^2y}{dx^2} = \left(2 \cdot 2e^{2x} \cos(3x) + 2e^{2x}(-3\sin(3x))\right) - \left(3 \cdot 2e^{2x} \sin(3x) + 3e^{2x}(3\cos(3x))\right) \]
Simplifying:
\[ = 4e^{2x} \cos(3x) - 6e^{2x} \sin(3x) - 6e^{2x} \sin(3x) - 9e^{2x} \cos(3x) \]
Combining like terms:
\[ = (4 - 9)e^{2x} \cos(3x) - 12e^{2x} \sin(3x) = -5e^{2x} \cos(3x) - 12e^{2x} \sin(3x) \]

### Part (iii): \( y = x^3 \log x \)

**Step 1: First Derivative**
Using the product rule:
\[ \frac{dy}{dx} = u'v + uv' \]
where \( u = x^3 \) and \( v = \log x \).

Calculating \( u' \) and \( v' \):
- \( u' = 3x^2 \)
- \( v' = \frac{1}{x} \)

Now applying the product rule:
\[ \frac{dy}{dx} = (3x^2)(\log x) + (x^3)\left(\frac{1}{x}\right) = 3x^2 \log x + x^2 \]

**Step 2: Second Derivative**
Now we differentiate \( \frac{dy}{dx} \):
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}(3x^2 \log x + x^2) \]
Using the product rule on \( 3x^2 \log x \):
\[ = 3(2x \log x + x) + 2x = 6x \log x + 3x + 2x = 6x \log x + 5x \]

### Final Answers:
1. \( \frac{d^2y}{dx^2} = 2\cos x - x \sin x \)
2. \( \frac{d^2y}{dx^2} = -5e^{2x} \cos(3x) - 12e^{2x} \sin(3x) \)
3. \( \frac{d^2y}{dx^2} = 6x \log x + 5x \)