Differentiate `(x+1)^(2)(x+2)^(3)(x+3)^(4)`

Explanation

To differentiate the expression \( (x+1)^2 (x+2)^3 (x+3)^4 \), we will use the product rule and the chain rule. Here’s a step-by-step solution:

### Step 1: Define the function
Let
\[ y = (x+1)^2 (x+2)^3 (x+3)^4 \]

### Step 2: Apply the logarithmic differentiation
Taking the natural logarithm of both sides:
\[ \ln y = \ln((x+1)^2) + \ln((x+2)^3) + \ln((x+3)^4) \]

### Step 3: Use logarithmic properties
Using the property of logarithms \(\ln(a^b) = b \ln a\):
\[ \ln y = 2 \ln(x+1) + 3 \ln(x+2) + 4 \ln(x+3) \]

### Step 4: Differentiate both sides
Now differentiate both sides with respect to \(x\):
\[ \frac{1}{y} \frac{dy}{dx} = 2 \cdot \frac{1}{x+1} \cdot \frac{d}{dx}(x+1) + 3 \cdot \frac{1}{x+2} \cdot \frac{d}{dx}(x+2) + 4 \cdot \frac{1}{x+3} \cdot \frac{d}{dx}(x+3) \]
Since \(\frac{d}{dx}(x+k) = 1\) for any constant \(k\), we have:
\[ \frac{1}{y} \frac{dy}{dx} = \frac{2}{x+1} + \frac{3}{x+2} + \frac{4}{x+3} \]

### Step 5: Solve for \(\frac{dy}{dx}\)
Multiplying both sides by \(y\):
\[ \frac{dy}{dx} = y \left( \frac{2}{x+1} + \frac{3}{x+2} + \frac{4}{x+3} \right) \]
Substituting back \(y = (x+1)^2 (x+2)^3 (x+3)^4\):
\[ \frac{dy}{dx} = (x+1)^2 (x+2)^3 (x+3)^4 \left( \frac{2}{x+1} + \frac{3}{x+2} + \frac{4}{x+3} \right) \]

### Step 6: Simplify the expression
To simplify:
\[ \frac{dy}{dx} = (x+1)^2 (x+2)^3 (x+3)^4 \left( \frac{2(x+2)(x+3) + 3(x+1)(x+3) + 4(x+1)(x+2)}{(x+1)(x+2)(x+3)} \right) \]

### Step 7: Combine terms
Now combine the terms in the numerator:
\[ = (x+1)^2 (x+2)^3 (x+3)^4 \left( \frac{2(x^2 + 5x + 6) + 3(x^2 + 4x + 3) + 4(x^2 + 3x + 2)}{(x+1)(x+2)(x+3)} \right) \]
Calculating the coefficients:
- \(2(x^2 + 5x + 6) = 2x^2 + 10x + 12\)
- \(3(x^2 + 4x + 3) = 3x^2 + 12x + 9\)
- \(4(x^2 + 3x + 2) = 4x^2 + 12x + 8\)

Combining:
\[ (2 + 3 + 4)x^2 + (10 + 12 + 12)x + (12 + 9 + 8) = 9x^2 + 34x + 29 \]

### Final expression
Thus, we have:
\[ \frac{dy}{dx} = (x+1)^2 (x+2)^3 (x+3)^4 \cdot \frac{9x^2 + 34x + 29}{(x+1)(x+2)(x+3)} \]