Differentiate w.r.t.x Let `y= (ax^(2) + bx + c)/( px^(2) + qx+ f)`

Answer

To differentiate the function \( y = \frac{ax^2 + bx + c}{px^2 + qx + f} \) with respect to \( x \), we will use the quotient rule of differentiation. The quotient rule states that if you have a function \( y = \frac{u}{v} \), then the derivative \( \frac{dy}{dx} \) is given by:

\[ \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]

### Step 1: Identify \( u \) and \( v \)

Let:
- \( u = ax^2 + bx + c \)
- \( v = px^2 + qx + f \)

### Step 2: Differentiate \( u \) and \( v \)

Now, we need to find \( \frac{du}{dx} \) and \( \frac{dv}{dx} \).

1. **Differentiate \( u \)**:
\[ \frac{du}{dx} = \frac{d}{dx}(ax^2 + bx + c) = 2ax + b \]

2. **Differentiate \( v \)**:
\[ \frac{dv}{dx} = \frac{d}{dx}(px^2 + qx + f) = 2px + q \]

### Step 3: Apply the Quotient Rule

Now, substitute \( u \), \( v \), \( \frac{du}{dx} \), and \( \frac{dv}{dx} \) into the quotient rule formula:

\[ \frac{dy}{dx} = \frac{(px^2 + qx + f)(2ax + b) - (ax^2 + bx + c)(2px + q)}{(px^2 + qx + f)^2} \]

### Step 4: Simplify the Expression

Now, we will simplify the numerator:

1. **Expand the first term**:
\[ (px^2 + qx + f)(2ax + b) = 2apx^3 + bpx^2 + 2aqx^2 + bqx + 2afx + bf \]

2. **Expand the second term**:
\[ (ax^2 + bx + c)(2px + q) = 2apx^3 + aqx^2 + 2pbx^2 + qbx + 2pcx + cq \]

3. **Combine both expansions**:
\[ \text{Numerator} = [2apx^3 + bpx^2 + 2aqx^2 + bqx + 2afx + bf] - [2apx^3 + aqx^2 + 2pbx^2 + qbx + 2pcx + cq] \]

4. **Cancel out similar terms**:
\[ \text{Numerator} = (bpx^2 - aqx^2 - 2pbx^2) + (2afx - 2pcx) + (bf - cq) \]
\[ = (bp - 2p - a)qx^2 + (2af - 2pc)x + (bf - cq) \]

### Final Result

Thus, the derivative \( \frac{dy}{dx} \) is:

\[ \frac{dy}{dx} = \frac{(bp - 2p - a)qx^2 + (2af - 2pc)x + (bf - cq)}{(px^2 + qx + f)^2} \]