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.
The key idea is the derivative at a point, defined as the limit of the difference quotient.
For f(x)=sinx, the derivative at x=0 is:
f′(0)=limh→0hsin(0+h)−sin0=limh→0hsinh
This is a standard limit: h→0limhsinh=1.
Therefore, the derivative of sinx at x=0 equals 1.
✓Final answer
The derivative is 1.
The derivative of sinx at x=0 is 1. This comes from the limit definition of the derivative and the fundamental limit limh→0hsinh=1.
The derivative of a function at a point tells us the slope of the tangent line at that point — the instantaneous rate of change. For sinx at x=0, we're asking: how fast is sinx changing exactly when x is zero?
If you picture the graph of sinx, it passes through the origin with a slope that looks like it might be 1 (since near x=0, sinx≈x). But let's prove it properly.
Start with the definition. The derivative of a function f(x) at a point x=a is:
Now evaluate the limit. This is the classic limit that defines the derivative of sine at zero. The key fact is:
limh→0hsinh=1
This is not obvious from plugging in h=0 (which gives 0/0), but it's a standard result proved using geometry or the squeeze theorem.
Tip
A quick intuition: for very small h, sinh≈h (in radians). So hsinh≈1, and the approximation gets better as h shrinks.
Therefore, the derivative is simply 1:
f′(0)=1
Watch out
A common mistake is to think the derivative of sinx is cosx everywhere, then plug in x=0 to get cos0=1. That's correct here, but only because we already know the general derivative formula. The limit definition is the foundation that justifies that formula.
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.