Q.Find the derivative of the function $f(x) = 2x^2 + 3x - 5$ at $x = -1$. Also prove that $f'(0) + 3f'(-1) = 0$.

Derivative at a Point: From Intuition to Precision

Imagine you're driving a car. Your speedometer doesn't tell you your average speed over the whole trip — it tells you your speed right now, at this exact instant. That's the core idea of a derivative at a point: it measures how fast something is changing at a single moment.

The Intuition: Instantaneous Rate of Change

Let's start with something simpler. Suppose you drop a ball from a height. The distance it has fallen after $t$ seconds is given by $s(t) = 4.9t^2$ metres (ignoring air resistance).

If I ask you "how fast was the ball falling after exactly 2 seconds?", you can't just divide distance by time — that gives an average speed over an interval. You need the speed at $t=2$, not between $t=1$ and $t=3$.

Here's the trick: take a very small time interval around $t=2$, say from $t=2$ to $t=2+h$ where $h$ is tiny. The average speed over that interval is:

If $h = 0.1$, you get one number. If $h = 0.01$, you get a slightly different number. As $h$ gets closer and closer to 0, these average speeds approach a single value — that's the instantaneous speed at $t=2$.

The Precise Definition

For a function $f(x)$, the derivative at a point $x = a$ is defined as:

provided this limit exists.

Let's break this down piece by piece:

• $f(a+h) - f(a)$ is the change in the function's value when you move from $a$ to $a+h$.
• Dividing by $h$ gives the average rate of change over that interval.
• Taking the limit as $h \to 0$ shrinks the interval to a single point, giving the instantaneous rate of change.

Geometric Interpretation

There's also a beautiful geometric meaning. The average rate of change $\frac{f(a+h) - f(a)}{h}$ is the slope of the secant line through the points $(a, f(a))$ and $(a+h, f(a+h))$.

As $h \to 0$, these two points get closer together, and the secant line approaches a line that just touches the curve at $x=a$ — the tangent line. So:

The derivative at a point equals the slope of the tangent line to the curve at that point.

A Concrete Example

Let's compute the derivative of $f(x) = x^2$ at $x = 3$.

Using the definition:

Expand $(3+h)^2 = 9 + 6h + h^2$:

Factor $h$:

Since $h \to 0$, this approaches $6$.

Notation

You'll see several notations for "the derivative of $f$ at $x=a$":

• $f'(a)$ — Lagrange notation (most common)
• $\frac{df}{dx}\big|_{x=a}$ — Leibniz notation
• $\dot{f}(a)$ — Newton notation (used mainly in physics for time derivatives)

All mean the same thing.

What If the Limit Doesn't Exist?

Not every function has a derivative at every point. The derivative fails to exist when:

• The function has a sharp corner (like $|x|$ at $x=0$)
• The function has a vertical tangent
• The function is discontinuous at that point

In such cases, we say the function is not differentiable at that point.

Why This Matters

The derivative at a point is the foundation of all of differential calculus. From it, you'll build:

• The derivative as a function (the derivative at every point)
• Rules for differentiation (product rule, chain rule, etc.)
• Applications: finding maxima/minima, related rates, curve sketching

But every single one of those starts here — with the idea of zooming in on a single point and asking: "How fast is this changing, right now?"

Derivative at a Point is introduced in the NCERT Class 11 Mathematics chapter on Limits and Derivatives and revisited in Class 12's Continuity and Differentiability, matching searches like "derivative definition using limits" or "differentiation important questions class 11 class 12 maths". This first-principles definition is a favourite CBSE board and JEE Main question type, since it tests genuine understanding rather than memorised differentiation rules.