Q.Find the derivative of the function 'cotx' from the first principle.

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:

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

The Precise Statement

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 $h = 0.1$:

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 $x = 3$.

Why You'll See It Everywhere

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