Q.Find the derivative of f(x)=1+x+x2+x3+⋯+x50 at x=1.
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Concept understanding — Derivative at a Point
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 — we differentiate term‑by‑term, then evaluate the resulting sum at x=1.
The function is a finite geometric series:
f(x)=1+x+x2+⋯+x50.
Differentiate term‑by‑term:
f′(x)=0+1+2x+3x2+⋯+50x49.
At x=1, this becomes the sum of the first 50 natural numbers:
f′(1)=1+2+3+⋯+50.
The sum of the first n natural numbers is 2n(n+1). Here n=50:
f′(1)=250×51=1275.
✓Final answer
The value is 1275.
The derivative of the sum f(x)=1+x+x2+⋯+x50 at x=1 is found by differentiating term-by-term and evaluating the resulting arithmetic series. The value is 1275.
The key insight: when you have a polynomial written as a sum of powers, the derivative is simply the sum of the derivatives of each term. At x=1, each derivative term nxn−1 becomes just n, so the problem reduces to adding the numbers from 1 to 50.
Differentiate term by term.
The derivative of a constant is 0. For each xn where n≥1, we have dxdxn=nxn−1. So:
f′(x)=0+1⋅x0+2⋅x1+3⋅x2+⋯+50⋅x49
That is:
f′(x)=1+2x+3x2+⋯+50x49
Evaluate at x=1.
At x=1, every power of x becomes 1. So:
f′(1)=1+2+3+⋯+50
Sum the arithmetic series.
The sum of the first n natural numbers is 2n(n+1). Here n=50:
f′(1)=250×51=25×51=1275
Watch out
A common mistake is to forget that the constant term 1 differentiates to 0, not to 1. Also, students sometimes stop at the expression 1+2+⋯+50 without actually summing it — in an exam, you must compute the numeric value.
Tip
If the series went up to x50, the derivative at x=1 is always the sum 1+2+⋯+50. For a general upper limit N, the answer is 2N(N+1). This pattern saves time in multiple-choice questions.