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

Explanation

To differentiate the function \( y = (x+1)^2 (x+2)^3 (x+3)^4 \) with respect to \( x \), we can use logarithmic differentiation for simplicity. Here are the steps:

### Step 1: Take the logarithm of both sides
Start by taking the natural logarithm of both sides:
\[ \ln y = \ln \left( (x+1)^2 (x+2)^3 (x+3)^4 \right) \]

### Step 2: Apply the properties of logarithms
Using the property of logarithms that states \( \ln(a \cdot b) = \ln a + \ln b \), we can expand the right-hand side:
\[ \ln y = \ln (x+1)^2 + \ln (x+2)^3 + \ln (x+3)^4 \]
Now applying the power rule of logarithms \( \ln(a^b) = b \ln a \):
\[ \ln y = 2 \ln (x+1) + 3 \ln (x+2) + 4 \ln (x+3) \]

### Step 3: Differentiate both sides
Now, differentiate both sides with respect to \( x \):
\[ \frac{d}{dx}(\ln y) = \frac{d}{dx}(2 \ln (x+1) + 3 \ln (x+2) + 4 \ln (x+3)) \]
Using the chain rule on the left side:
\[ \frac{1}{y} \frac{dy}{dx} = 2 \cdot \frac{1}{x+1} \cdot \frac{d}{dx}(x+1) + 3 \cdot \frac{1}{x+2} \cdot \frac{d}{dx}(x+2) + 4 \cdot \frac{1}{x+3} \cdot \frac{d}{dx}(x+3) \]
This simplifies to:
\[ \frac{1}{y} \frac{dy}{dx} = \frac{2}{x+1} + \frac{3}{x+2} + \frac{4}{x+3} \]

### Step 4: Solve for \( \frac{dy}{dx} \)
Now, multiply both sides by \( y \) to isolate \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = y \left( \frac{2}{x+1} + \frac{3}{x+2} + \frac{4}{x+3} \right) \]
Substituting back the expression for \( y \):
\[ \frac{dy}{dx} = (x+1)^2 (x+2)^3 (x+3)^4 \left( \frac{2}{x+1} + \frac{3}{x+2} + \frac{4}{x+3} \right) \]

### Final Result
Thus, the derivative of the function \( y = (x+1)^2 (x+2)^3 (x+3)^4 \) with respect to \( x \) is:
\[ \frac{dy}{dx} = (x+1)^2 (x+2)^3 (x+3)^4 \left( \frac{2}{x+1} + \frac{3}{x+2} + \frac{4}{x+3} \right) \]