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In Class 11 you learn rules such as "the derivative of xⁿ is nxⁿ⁻¹" in a single lesson. Before that, NCERT asks you to find derivatives the long way, from the definition. Questions that say "from first principles" expect this method, and students lose marks not because the algebra is hard but because they skip steps or let h become zero too early. This post explains what the definition means, works through five examples, and shows the slips to avoid.
Where is this in the syllabus?
It is Chapter 12, Limits and Derivatives, of the NCERT Class 11 mathematics textbook. The CBSE 2026-27 mathematics curriculum lists "Definition of derivative relate it to scope of tangent of the curve, derivative of sum, difference, product and quotient of functions of polynomial and trigonometric functions". The JEE (Main) 2026 syllabus includes "limits, continuity and differentiability". Check the current year's documents before you plan.
What does "first principles" mean?
NCERT's Definition 2 says the derivative of f at x is
f′(x) = lim (h → 0) [f(x + h) − f(x)] / h
"wherever the limit exists", and adds: "This definition of derivative is also called the first principle of derivative."
The fraction is the slope of the line joining two points on the graph, at x and at x + h. As h shrinks, that line turns into the tangent at x, and its slope becomes the derivative.
Worked example 1: watch the limit happen
Take f(x) = x² at x = 3.
h = 0.1: (3.1² − 9) / 0.1 = (9.61 − 9) / 0.1 = 6.1
h = 0.01: (3.01² − 9) / 0.01 = (9.0601 − 9) / 0.01 = 6.01
The values approach 6, which is 2x at x = 3. The definition turns this pattern into algebra.
Worked example 2: f(x) = x³
f(x + h) − f(x) = (x + h)³ − x³ = 3x²h + 3xh² + h³
Divide by h: 3x² + 3xh + h²
Let h → 0: f′(x) = 3x²
The order matters. You cannot put h = 0 in step 1, because the fraction would be 0/0. You cancel the h first, then take the limit.
Worked example 3: f(x) = √x, for x > 0
Here the trick is to rationalise.
[√(x + h) − √x] / h × [√(x + h) + √x] / [√(x + h) + √x] = (x + h − x) / [h(√(x + h) + √x)] = 1 / (√(x + h) + √x)
Now let h → 0: f′(x) = 1 / (2√x).
At x = 0 the same fraction becomes √h / h = 1 / √h, which grows without limit as h → 0 from the right, so √x has no derivative at 0. The graph has a vertical tangent there.
Worked example 4: f(x) = sin x
Use sin(x + h) = sin x cos h + cos x sin h:
[sin(x + h) − sin x] / h = cos x × (sin h / h) − sin x × (1 − cos h) / h
NCERT's Theorem 5 in the same chapter proves the two limits you need: sin h / h → 1 and (1 − cos h) / h → 0 as h → 0. So f′(x) = cos x × 1 − sin x × 0 = cos x.
Worked example 5: |x| at x = 0
f(0 + h) − f(0) = |h|, so the fraction is |h| / h.
From the right (h > 0): |h| / h = 1
From the left (h < 0): |h| / h = −1
The one-sided limits are different, so the limit does not exist and |x| has no derivative at 0, even though it is continuous there. The graph has a corner. Our post on limits and continuity covers one-sided limits in more depth.
Why does the definition confuse so many students?
Two lines of research help. In clinical interviews with 110 students, Adrian Orton (1983) collected detailed data on how well they understood differentiation and rate of change, and on their common errors. A later study by Jungeun Park (2015) watched three calculus instructors introduce the derivative and found that they moved from the derivative at a point to the derivative as a function mainly by swapping the letter for a point for a variable, without discussing how the quantity changes over an interval.
That second finding matches the jump in this chapter. Example 1 finds a number (6 at x = 3). Examples 2 to 4 find a new function (3x², 1/(2√x), cos x). Say which one a question asks for before you start.
Which mistakes cost marks?
Expanding (x + h)² as x² + h². The middle term 2xh is the derivative.
Putting h = 0 too early, before cancelling, and writing 0/0 = 1.
Dropping "lim" from the lines in between. Examiners give method marks for the notation.
Using the power rule when the question says "from first principles". You will lose most of the marks.
Forgetting the domain. √x works only for x > 0 in this calculation.
If you keep getting stuck at the simplifying step, the Fermi Problem Library has a step-by-step solver that explains the reasoning rather than printing an answer, so you can see which identity or rationalisation the next line needs.
A practice routine
Derive x², x³, 1/x, √x, sin x and cos x from first principles, once each, in the three-line pattern.
For two of them, make an h-table like Example 1 at a point and check it approaches your formula.
Do NCERT Exercise 12.2 with the book closed, then compare with the rules from the rest of the chapter.
Next, see how these derivatives become slopes of real lines in tangents and normals.
Tonight, find the derivative of cos x from first principles and check it against −sin x.
