Differentiate with respect to $x$ using first principle: $\dfrac{ax + b}{cx + d}$.

> [!TLDR] > The derivative of $\frac{ax+b}{cx+d}$ is found using the first principle of differentiation (limit definition). The final result is $\frac{ad - bc}{(cx+d)^2}$. The first principle — also called the definition of the derivative — asks us to compute the limit of the difference quotient as the change in $x$ approaches zero. For a function $f(x)$, this is: $$ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}. $$ Here, $f(x) = \frac{ax+b}{cx+d}$. The key challenge is algebraic: we need to combine the two fractions in the numerator, simplify, and then take the limit. The quotient rule is a shortcut for this, but using first principles directly builds a deeper understanding of why the derivative formula works. Let’s go step by step. 1. **Write the difference quotient.** We need: $$ \frac{f(x+h) - f(x)}{h} = \frac{ \frac{a(x+h)+b}{c(x+h)+d} - \frac{ax+b}{cx+d} }{h}. $$ 2. **Combine the two fractions in the numerator.** The common denominator is $(c(x+h)+d)(cx+d)$. So: $$ \frac{a(x+h)+b}{c(x+h)+d} - \frac{ax+b}{cx+d} = \frac{ [a(x+h)+b](cx+d) - (ax+b)[c(x+h)+d] }{ (c(x+h)+d)(cx+d) }. $$ 3. **Expand the numerator carefully.** First term: $[a(x+h)+b](cx+d) = a(x+h)(cx+d) + b(cx+d)$. Second term: $(ax+b)[c(x+h)+d] = (ax+b)[c(x+h)] + (ax+b)d$. It’s easier to expand systematically: - $[a(x+h)+b](cx+d) = a(x+h)(cx) + a(x+h)d + b(cx) + bd$ $= a c x(x+h) + a d (x+h) + b c x + b d$. - $(ax+b)[c(x+h)+d] = ax \cdot c(x+h) + ax \cdot d + b \cdot c(x+h) + b \cdot d$ $= a c x (x+h) + a d x + b c (x+h) + b d$. Now subtract the second from the first: $$ \text{Numerator} = [a c x(x+h) + a d (x+h) + b c x + b d] - [a c x(x+h) + a d x + b c (x+h) + b d]. $$ Notice $a c x(x+h)$ cancels, and $b d$ cancels. We are left with: $$ a d (x+h) - a d x + b c x - b c (x+h) = a d h + b c (x - (x+h)) = a d h - b c h = h(ad - bc). $$ …

Differentiate with respect to x using first principle: (ax + b)/(cx + d).

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Concept understanding — Difference Quotient

The Difference Quotient: What It Is and Why It Matters

Imagine you're tracking the distance a car has travelled over time. At 2:00 PM, the odometer reads 40 km. At 2:30 PM, it reads 70 km. How fast was the car going on average during that half-hour?

You'd calculate: $\frac{70 - 40}{0.5 \text{ hours}} = 60$ km/h.

That fraction — change in distance divided by change in time — is the average rate of change. The difference quotient is just a formal, algebraic way of writing that same idea for any function.

The Intuition: Slope of a Secant Line

Take any function $f(x)$ . Pick two points on its graph: $(x, f(x))$ and $(x + h, f(x + h))$ , where $h$ is some horizontal step (positive or negative). The line that cuts through both points is called a secant line.

The slope of that secant line is:

\text{slope} = \frac{\text{rise}}{\text{run}} = \frac{f(x + h) - f(x)}{(x + h) - x} = \frac{f(x + h) - f(x)}{h}

That expression — $\frac{f(x + h) - f(x)}{h}$ — is the difference quotient.

Note

The name comes from "difference" (you subtract two function values) and "quotient" (you divide by $h$ ). It's literally a quotient of differences.

The Precise Statement

\frac{f(x + h) - f(x)}{h}, \quad h \neq 0

This gives the average rate of change of $f$ over the interval from $x$ to $x + h$ . Geometrically, it's the slope of the secant line through $(x, f(x))$ and $(x + h, f(x + h))$ .

Key restrictions:

$h$ cannot be zero (you can't divide by zero).
$x$ and $x + h$ must both be in the domain of $f$ .

A Concrete Example

Let $f(x) = x^2$ . Compute the difference quotient at $x = 3$ with $h = 0.1$ :

\frac{f(3 + 0.1) - f(3)}{0.1} = \frac{(3.1)^2 - 9}{0.1} = \frac{9.61 - 9}{0.1} = \frac{0.61}{0.1} = 6.1

This tells us: over the interval $[3, 3.1]$ , the function $x^2$ increases at an average rate of 6.1 units per unit change in $x$ .

If you shrink $h$ to 0.01, you'd get 6.01. As $h$ gets smaller, the average rate approaches 6 — which is exactly the instantaneous rate of change (the derivative) of $x^2$ at $x = 3$ .

Tip

The difference quotient is the bridge between average rates (which you can compute with simple algebra) and instantaneous rates (which require limits). When you take the limit as $h \to 0$ , you get the derivative.

Why You'll See It Everywhere

The difference quotient isn't just a classroom exercise. It's the foundation of calculus: …

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The derivative of $\frac{ax+b}{cx+d}$ is found using the first principle of differentiation (limit definition). The final result is $\frac{ad - bc}{(cx+d)^2}$ .

The first principle — also called the definition of the derivative — asks us to compute the limit of the difference quotient as the change in $x$ approaches zero. For a function $f(x)$ , this is:

f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}.

Here, $f(x) = \frac{ax+b}{cx+d}$ . The key challenge is algebraic: we need to combine the two fractions in the numerator, simplify, and then take the limit. The quotient rule is a shortcut for this, but using first principles directly builds a deeper understanding of why the derivative formula works.

Let’s go step by step.

Write the difference quotient. We need:

\frac{f(x+h) - f(x)}{h} = \frac{ \frac{a(x+h)+b}{c(x+h)+d} - \frac{ax+b}{cx+d} }{h}.

Combine the two fractions in the numerator. The common denominator is $(c(x+h)+d)(cx+d)$ . So:

\frac{a(x+h)+b}{c(x+h)+d} - \frac{ax+b}{cx+d} = \frac{ [a(x+h)+b](cx+d) - (ax+b)[c(x+h)+d] }{ (c(x+h)+d)(cx+d) }.

Expand the numerator carefully.

First term: $[a(x+h)+b](cx+d) = a(x+h)(cx+d) + b(cx+d)$ .

Second term: $(ax+b)[c(x+h)+d] = (ax+b)[c(x+h)] + (ax+b)d$ .

It’s easier to expand systematically:
- $[a(x+h)+b](cx+d) = a(x+h)(cx) + a(x+h)d + b(cx) + bd$ $= a c x(x+h) + a d (x+h) + b c x + b d$ .
- $(ax+b)[c(x+h)+d] = ax \cdot c(x+h) + ax \cdot d + b \cdot c(x+h) + b \cdot d$ $= a c x (x+h) + a d x + b c (x+h) + b d$ .
Now subtract the second from the first:

\text{Numerator} = [a c x(x+h) + a d (x+h) + b c x + b d] - [a c x(x+h) + a d x + b c (x+h) + b d].

Notice $a c x(x+h)$ cancels, and $b d$ cancels. We are left with:

$a d (x+h) - a d x + b c x - b c (x+h) = a d h + b c (x - (x+h)) = a d h - b c h = h(ad - bc).$ …

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