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)=4.9t2 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 att=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:
hs(2+h)−s(2)
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.
Note
This "shrinking interval" idea is the heart of the derivative. We're not setting h=0 (that would give 00, which is meaningless). We're letting happroach 0 and seeing what the ratio approaches.
The Precise Definition
For a function f(x), the derivative at a pointx=a is defined as:
f′(a)=limh→0hf(a+h)−f(a)
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→0 shrinks the interval to a single point, giving the instantaneous rate of change.
f′(a)=limh→0hf(a+h)−f(a)
Geometric Interpretation
There's also a beautiful geometric meaning. The average rate of change hf(a+h)−f(a) is the slope of the secant line through the points (a,f(a)) and (a+h,f(a+h)).
As h→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)=x2 at x=3.
Using the definition:
f′(3)=limh→0h(3+h)2−32
Expand (3+h)2=9+6h+h2:
f′(3)=limh→0h9+6h+h2−9=limh→0h6h+h2
Factor h:
f′(3)=limh→0hh(6+h)=limh→0(6+h)
Since h→0, this approaches 6.
Important
The derivative of x2 at x=3 is 6. This means:
At x=3, the function is increasing at a rate of 6 units per unit change in x.
The tangent line to y=x2 at (3,9) has slope 6.
Notation
You'll see several notations for "the derivative of f at x=a":
f′(a) — Lagrange notation (most common)
dxdfx=a — Leibniz notation
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.
Concept: Derivative at a Point — the slope of the tangent line, found using the limit definition or a differentiation rule.
Step 1: Let f(x)=x2−2. The derivative f′(x) is found by the power rule:
dxd(x2)=2x, and the derivative of a constant is 0.
Step 2: So f′(x)=2x.
Step 3: Evaluate at x=10:
f′(10)=2(10)=20.
✓Final answer
The derivative at x=10 is 20.
The derivative of x2−2 at x=10 is 20. We find this by first computing the general derivative 2x using the power rule, then evaluating at x=10.
The derivative of a function at a specific point tells us the instantaneous rate of change — the slope of the tangent line — at that exact location. For a polynomial like x2−2, the derivative is straightforward to compute, and then we simply plug in the given x-value.
Why this works: The derivative of xn is nxn−1 (the power rule), and the derivative of a constant is 0. So the derivative of x2−2 is 2x−0=2x. That’s the general formula for the slope at any point x. To get the slope at x=10, we substitute 10 into 2x.
Let’s go through it step by step.
Identify the function.
We have f(x)=x2−2. This is a simple quadratic with a constant term.
Apply the power rule to the x2 term.
The power rule says: if f(x)=xn, then f′(x)=nxn−1.
Here n=2, so the derivative of x2 is 2x2−1=2x.
Handle the constant term.
The derivative of any constant (like −2) is 0, because a constant doesn’t change — its rate of change is zero.
Combine the results.
So f′(x)=2x+0=2x.
Evaluate at x=10.
Substitute 10 into the derivative: f′(10)=2×10=20.
Watch out
A common mistake is to forget that the derivative of a constant is zero, or to incorrectly apply the power rule to the constant term. Also, some students try to plug x=10 into the original function first and then differentiate — that doesn’t work because the derivative is about the function’s behavior around a point, not just at it.
Tip
If you ever need the derivative at a point quickly for a polynomial, you can often do it in one step: bring down the exponent, reduce it by one, and then evaluate. For x2, that’s 2x, then 2(10)=20.
Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.
CBSE 2026Set 1A1 mark
Q.Find the derivative of x2−2 at x=10.
›Reveal solutionSolution
dxd(x2−2)=2x, which is 20 at x=10.
If f(x)=x2−2 then f′(x)=2x. At x=10, f′(10)=2(10)=20.
✓Final answer
f′(10)=20.
CBSE 2025Set ANNUAL1 markMCQ
Q.limx→10[dxd(x2−2)]=
(a) 10
(b) 20
(c) 50
(d) 100
›Reveal solutionSolution
limx→10[dxd(x2−2)]=20.
First find the derivative: dxd(x2−2)=2x−0=2x.
Now take the limit of this derivative as x→10: since 2x is continuous, substitute x=10: 2(10)=20.
✓Final answer
(b) 20.
CBSE 2024Set ANNUAL1 markMCQ
Q.If f(x)=ax2+bx+c, then f′(0)=
(a) b
(b) a
(c) c
(d) None of these
›Reveal solutionSolution
Differentiate term by term using the power rule, then substitute x=0.
Given f(x)=ax2+bx+c, differentiate each term:
dxd(ax2)=2ax
dxd(bx)=b
dxd(c)=0 (constant)
So:
f′(x)=2ax+b
Substitute x=0:
f′(0)=2a(0)+b=b
✓Final answer
(a) b.
CBSE 2023Set ANNUAL1 markMCQ
Q.The derivative of 99x at x=100 will be:
(a) 100
(b) 99x
(c) 99
(d) 9900
›Reveal solutionSolution
dxd(99x)=99 for every x, including x=100.
For a linear function f(x)=cx (with constant c), the derivative is simply f′(x)=c for every x, since the slope of a straight line is the same everywhere.
Here f(x)=99x, so f′(x)=99 for all x. In particular f′(100)=99 — the specific point x=100 does not matter, since the derivative of a linear function is constant.
✓Final answer
The correct option is (c) 99.
CBSE 2023Set ANNUAL1 mark
Q.Fill in the blank: The derivative of f(x)=99x at x=100 is ____.
›Reveal solutionSolution
The derivative of f(x)=99x is the constant 99, at every point including x=100.
For f(x)=99x, using the standard rule dxd(cx)=c:
f′(x)=99
This derivative does not depend on x, so at x=100, f′(100)=99.
✓Final answer
f′(100)=99.
CBSE 2022Set TERM11 markMCQ
Q.If f(x)=2x2+3x−5 then f′(0)+3f′(−1)=
(a) 2
(b) 0
(c) 1
(d) −1
›Reveal solutionSolution
Differentiate first, then substitute the required x-values.
f(x)=2x2+3x−5⇒f′(x)=4x+3. f′(0)=4(0)+3=3. f′(−1)=4(−1)+3=−1. So f′(0)+3f′(−1)=3+3(−1)=3−3=0.