`((5x^(2) + 6x+ 7))/((2x^(2) + 3x +4))`

Answer

To differentiate the function \( \frac{5x^2 + 6x + 7}{2x^2 + 3x + 4} \) with respect to \( x \), we will use the quotient rule. The quotient rule states that if you have a function in the form \( \frac{U}{V} \), then the derivative is given by:

\[ \frac{d}{dx}\left(\frac{U}{V}\right) = \frac{V \cdot \frac{dU}{dx} - U \cdot \frac{dV}{dx}}{V^2} \]

### Step 1: Identify \( U \) and \( V \)
Let:
- \( U = 5x^2 + 6x + 7 \)
- \( V = 2x^2 + 3x + 4 \)

### Step 2: Differentiate \( U \) and \( V \)
Now we differentiate \( U \) and \( V \):
- \( \frac{dU}{dx} = 10x + 6 \)
- \( \frac{dV}{dx} = 4x + 3 \)

### Step 3: Apply the Quotient Rule
Using the quotient rule, we have:

\[ \frac{d}{dx}\left(\frac{U}{V}\right) = \frac{V \cdot \frac{dU}{dx} - U \cdot \frac{dV}{dx}}{V^2} \]

Substituting \( U, V, \frac{dU}{dx}, \) and \( \frac{dV}{dx} \):

\[ \frac{d}{dx}\left(\frac{5x^2 + 6x + 7}{2x^2 + 3x + 4}\right) = \frac{(2x^2 + 3x + 4)(10x + 6) - (5x^2 + 6x + 7)(4x + 3)}{(2x^2 + 3x + 4)^2} \]

### Step 4: Simplify the Numerator
Now we simplify the numerator:

1. Expand \( (2x^2 + 3x + 4)(10x + 6) \):
\[ = 20x^3 + 12x^2 + 30x^2 + 18x + 40x + 24 = 20x^3 + 42x^2 + 58x + 24 \]

2. Expand \( (5x^2 + 6x + 7)(4x + 3) \):
\[ = 20x^3 + 15x^2 + 24x + 21 \]

3. Combine these results:
\[ \text{Numerator} = (20x^3 + 42x^2 + 58x + 24) - (20x^3 + 15x^2 + 24x + 21) \]
\[ = 20x^3 + 42x^2 + 58x + 24 - 20x^3 - 15x^2 - 24x - 21 \]
\[ = (42x^2 - 15x^2) + (58x - 24x) + (24 - 21) \]
\[ = 27x^2 + 34x + 3 \]

### Step 5: Write the Final Result
Thus, the derivative is:

\[ \frac{d}{dx}\left(\frac{5x^2 + 6x + 7}{2x^2 + 3x + 4}\right) = \frac{27x^2 + 34x + 3}{(2x^2 + 3x + 4)^2} \]