Find the derivative of $\dfrac{px^2 + qx + r}{ax + b}$.

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> [!TLDR]
> Apply the Quotient Rule to differentiate a rational function: the derivative of $\frac{u}{v}$ is $\frac{v \cdot u' - u \cdot v'}{v^2}$. The result is **$\dfrac{apx^2 + 2bpx + bq - ar}{(ax+b)^2}$**.

When you have one polynomial divided by another, the Quotient Rule is your tool. The idea is simple: if $y = \frac{u(x)}{v(x)}$, then the rate of change of $y$ depends on how fast the numerator is changing *relative to* the denominator, adjusted for how the denominator itself is changing. The formula captures this interplay:

> [!FORMULA]
> $$\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2}$$

The denominator $v^2$ appears because we're measuring the rate per unit of the "base" (the denominator), and both terms in the numerator account for the coupled changes.

Let me work through this problem step by step.

**1. Identify the numerator and denominator**

Here $u(x) = px^2 + qx + r$ and $v(x) = ax + b$.

**2. Differentiate the numerator**

$$\frac{du}{dx} = \frac{d}{dx}(px^2 + qx + r) = 2px + q$$

**3. Differentiate the denominator**

$$\frac{dv}{dx} = \frac{d}{dx}(ax + b) = a$$

**4. Apply the Quotient Rule**

Substitute into the formula:

$$\frac{d}{dx}\left(\frac{px^2 + qx + r}{ax + b}\right) = \frac{(ax + b)(2px + q) - (px^2 + qx + r)(a)}{(ax + b)^2}$$

**5. Expand the numerator**

First term: $(ax + b)(2px + q)$

$$= ax \cdot 2px + ax \cdot q + b \cdot 2px + b \cdot q = 2apx^2 + aqx + 2bpx + bq$$

Second term: $(px^2 + qx + r) \cdot a$

$$= apx^2 + aqx + ar$$ …

Q.Find the derivative of $\dfrac{px^2 + qx + r}{ax + b}$ .

Concept understanding — Rules of Differentiation

Q.Find the derivative of px2+qx+rax+b\dfrac{px^2 + qx + r}{ax + b}ax+bpx2+qx+r​.

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Q.Find the derivative of $\dfrac{px^2 + qx + r}{ax + b}$ .

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