MathsCBSE Class 11 · Limits and Derivatives

Derivatives: First Principle, Product and Quotient Rules (Class 11)

Updated 17 October 2026 · Fact-checked

The derivative of f at x is f'(x) = lim (f(x + h) − f(x))/h as h tends to 0. Standard results: d/dx (x^n) = n x^(n−1), d/dx (sin x) = cos x, d/dx (cos x) = −sin x. Product rule: (uv)' = u'v + uv'. Quotient rule: (u/v)' = (u'v − uv')/v². The derivative at a point gives the slope of the tangent and the instantaneous rate of change.

Understand Derivatives: First Principle and Rules of Differentiation

The derivative of $y = f (x)$ is
$ $f^{'} (x) = lim_{h \to 0} \frac{f ( x + h ) - f ( x )}{h},$ $
the limit of the average rate of change over a shrinking interval. Geometrically it is the slope of the tangent at $(x, f (x))$ ; physically it is the instantaneous rate of change.

First principle. To differentiate from the definition: write $f (x + h) - f (x)$ , simplify (often by rationalising or using an identity), divide by $h$ , cancel the $h$ , then let $h \to 0$ . For example, $f (x) = \frac{1}{x}$ :
$ $\frac{\frac{1}{x + h} - \frac{1}{x}}{h} = \frac{- h}{h x ( x + h )} = \frac{- 1}{x ( x + h )} \to - \frac{1}{x ^{2}} .$ $
For $sin x$ , use $sin (x + h) - sin x = 2 cos (x + \frac{h}{2}) sin \frac{h}{2}$ ; the limit $\frac{s i n ( h /2 )}{h /2} \to 1$ gives $cos x$ .

**Rules (for differentiable $u, v$ ):**

Sum: $(u \pm v)^{'} = u^{'} \pm v^{'}$ ; constant multiple: $(c u)^{'} = c u^{'}$ ; derivative of a constant is $0$ .
Power: $\frac{d}{d x} x^{n} = n x^{n - 1}$ .
Product: $(uv)^{'} = u^{'} v + u v^{'}$ .
Quotient: $(\frac{u}{v})^{'} = \frac{u ^{'} v - u v ^{'}}{v ^{2}}$ , $v \neq = 0$ .
Trigonometric: $(sin x)^{'} = cos x$ , $(cos x)^{'} = - sin x$ , $(tan x)^{'} = sec^{2} x$ , $(sec x)^{'} = sec x tan x$ , and similarly for $cot x$ and $cosec x$ .

The chain rule for composite functions is only formative in 2026-27, so questions here use these rules directly.

Tangent and rate of change. The slope of the tangent at $x = a$ is $f^{'} (a)$ , and the tangent is $y - f (a) = f^{'} (a) (x - a)$ . A tangent parallel to the $x$ -axis has $f^{'} (x) = 0$ ; a tangent parallel to $y = m x + c$ has $f^{'} (x) = m$ . If $s (t)$ is position, velocity is $s^{'} (t)$ and acceleration is the derivative of the velocity.

Simplify before differentiating when you can: expanding $(x^{2} + 1) (x^{3} - 2 x)$ and using the power rule avoids the product rule, and an identity can turn a quotient into a simple expression.

Derivatives: First Principle and Rules of Differentiation: formulas to remember

Definition: $f^{'} (x) = l im (f (x + h) - f (x)) / ha s h \to 0$; First principle.
Power rule: $d / d x (x^{n}) = n x^{(} n - 1)$; Any real n.
Product rule: $(uv)^{'} = u^{'} v + u v^{'}$; Not u'v'.
Quotient rule: $(u / v)^{'} = (u^{'} v - u v^{'}) / v^{2}$; Order of terms in the numerator matters.
Trigonometric: $(s in x)^{'} = cos x; (cos x)^{'} = - s in x; (t an x)^{'} = se c^{2} x$; Also (sec x)' = sec x tan x.
Tangent: $y - f (a) = f^{'} (a) (x - a)$; Slope = f'(a).

How to solve Derivatives: First Principle and Rules of Differentiation questions

Use these steps for first-principle proofs, rule-based differentiation and tangent problems.

1Decide whether the question demands first principle; if it does, write the difference quotient and simplify before letting $h \to 0$ .
2Otherwise, identify the structure: sum, product or quotient, and choose the rule; first see whether expanding or an identity makes it simpler.
3Label $u$ and $v$ and write $u^{'}$ and $v^{'}$ separately before combining.
4Combine using the rule, then simplify the result (factor, use identities).
5For a tangent, find $f^{'} (a)$ , evaluate $f (a)$ and write the line equation.
6For motion, differentiate once for velocity and again for acceleration; substitute the times asked.

Derivatives: First Principle and Rules of Differentiation in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Rate of change of area

Suppose a square tile expands evenly while it is baked, so its side $x$ cm grows, and its area is $A = x^{2}$ .

How fast does the area change per cm of side when $x = 5$ cm?

Try it first, then show the solutionSolution

$\frac{d A}{d x} = 2 x$ .
At $x = 5$ : $2 \times 5 = 10$ cm² per cm.
Check: from $x = 5$ to $5.01$ , area goes from $25$ to $25.1001$ , a change of $0.1001$ for $0.01$ cm: about $10$ per cm.

Answer: 10 cm² per cm of side.

Memory hook: The derivative is the exchange rate between two quantities.

Slope of a ramp

Suppose a skate ramp has the profile $y = \frac{x ^{2}}{20}$ (both in metres).

Find the slope of the ramp at $x = 10$ m and the angle it makes with the horizontal.

Try it first, then show the solutionSolution

$\frac{d y}{d x} = \frac{x}{10}$ .
At $x = 10$ : slope $= 1$ .
Slope $= tan θ = 1$ , so $θ = 4 5^{\circ}$ .

Answer: Slope 1; angle $4 5^{\circ}$ .

Memory hook: The derivative is the slope; slope is tan of the angle.

Tricks to remember Derivatives: First Principle and Rules of Differentiation

Short memory aids. Say each one aloud once, then use it on a question the same day.

Product rule sentence
Derivative of the first times the second, plus the first times the derivative of the second.
Use it when: Product rule.
Lo d-hi minus hi d-lo
For the quotient: (bottom × derivative of top − top × derivative of bottom) over bottom squared.
Use it when: Quotient rule order.
Expand first
If a product of polynomials is small, multiply it out and use the power rule.
Use it when: Saves marks lost to errors.
Zero slope
Horizontal tangent means f'(x) = 0; parallel to y = mx + c means f'(x) = m.
Use it when: Tangent questions.

Derivatives: First Principle and Rules of Differentiation: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: First principle · 3 marks · Short answerBoard level

Find the derivative of $f (x) = x$ ( $x > 0$ ) from first principle.

Try it first, then show the solutionSolution

$f^{'} (x) = lim_{h \to 0} \frac{x + h - x}{h}$ . (1 mark)
Multiply by $x + h + x$ : numerator becomes $(x + h) - x = h$ . (1 mark)
Cancel $h$ : $\frac{1}{x + h + x} \to \frac{1}{2 x}$ as $h \to 0$ . (1 mark)

Answer: $\frac{1}{2 x}$

Example 2: Product rule · 2 marks · Very short answerBoard level

Differentiate $(x^{2} + 1) (x^{3} - 2 x)$ and check by expanding.

Try it first, then show the solutionSolution

$u = x^{2} + 1$ , $v = x^{3} - 2 x$ ; $u^{'} v + u v^{'} = 2 x (x^{3} - 2 x) + (x^{2} + 1) (3 x^{2} - 2) = 5 x^{4} - 3 x^{2} - 2$ . (1 mark)
Check: $(x^{2} + 1) (x^{3} - 2 x) = x^{5} - x^{3} - 2 x$ , whose derivative is $5 x^{4} - 3 x^{2} - 2$ . (1 mark)

Answer: $5 x^{4} - 3 x^{2} - 2$

Example 3: Quotient with an identity · 3 marks · Short answerHarder board level

Differentiate $\frac{cos x}{1 + sin x}$ and simplify.

Try it first, then show the solutionSolution

$\frac{( - sin x ) ( 1 + sin x ) - cos x \cdot cos x}{( 1 + sin x ) ^{2}}$ . (1 mark)
Numerator $= - sin x - sin^{2} x - cos^{2} x = - (1 + sin x)$ . (1 mark)
Result $= \frac{- ( 1 + sin x )}{( 1 + sin x ) ^{2}} = - \frac{1}{1 + sin x}$ . (1 mark)

Answer: $- \frac{1}{1 + sin x}$

Example 4: Tangent to a curve · 3 marks · Short answerHarder board level

Find the equation of the tangent to $y = x sin x$ at $x = \frac{π}{2}$ .

Try it first, then show the solutionSolution

$\frac{d y}{d x} = sin x + x cos x$ (product rule). (1 mark)
At $x = \frac{π}{2}$ : slope $= 1 + \frac{π}{2} \cdot 0 = 1$ ; the point is $(\frac{π}{2}, \frac{π}{2})$ . (1 mark)
Tangent: $y - \frac{π}{2} = 1 (x - \frac{π}{2})$ , i.e. $y = x$ . (1 mark)

Answer: $y = x$

Example 5: Case study: motion in a straight line · 4 marks · Case-basedHOTS / competency

Suppose a robot moves along a rail so that its position is $s = 2 t^{3} - 15 t^{2} + 36 t$ metres at time $t$ seconds, $t \geq 0$ .
(i) Find its velocity. (1 mark)
(ii) At what times is it momentarily at rest? (1 mark)
(iii) Find its position at those times. (1 mark)
(iv) Find its acceleration at those times. (1 mark)

Try it first, then show the solutionSolution

(i) $v = \frac{d s}{d t} = 6 t^{2} - 30 t + 36 = 6 (t - 2) (t - 3)$ . (1 mark)
(ii) $v = 0$ at $t = 2$ and $t = 3$ . (1 mark)
(iii) $s (2) = 16 - 60 + 72 = 28$ m; $s (3) = 54 - 135 + 108 = 27$ m. (1 mark)
(iv) $a = \frac{d v}{d t} = 12 t - 30$ : $a (2) = - 6$ m/s², $a (3) = 6$ m/s². (1 mark)

Answer: (i) $6 t^{2} - 30 t + 36$ (ii) $t = 2, 3$ (iii) 28 m and 27 m (iv) $- 6$ and $6$ m/s²

Where students lose marks in Derivatives: First Principle and Rules of Differentiation

The slips examiners see most often, and what to write instead.

Mistake: Product rule as u'v'
Students differentiate each factor and multiply.
Wrong working
$(x^{2} sin x)^{'} = 2 x cos x$ .
Right working
$2 x sin x + x^{2} cos x$ .
Do this: Say "first times derivative of second, plus second times derivative of first".
Usually costs: The whole answer.
Mistake: Reversing the quotient rule numerator
The order u'v − uv' matters.
Wrong working
$(\frac{x}{x + 1})^{'} = \frac{x - ( x + 1 )}{( x + 1 ) ^{2}}$ .
Right working
$\frac{1 \cdot ( x + 1 ) - x \cdot 1}{( x + 1 ) ^{2}} = \frac{1}{( x + 1 ) ^{2}}$ .
Do this: Write u, v, u' and v' first.
Usually costs: A sign error costs the answer.
Mistake: Dropping h too early in first principle
Setting h = 0 before cancelling gives 0/0.
Wrong working
$\frac{\frac{1}{x + h} - \frac{1}{x}}{h}$ with $h = 0$ .
Right working
Simplify to $\frac{- 1}{x ( x + h )}$ first, then let $h \to 0$ .
Do this: Cancel h before taking the limit.
Usually costs: Method marks.
Mistake: Wrong sign for the derivative of cos x
The negative sign is forgotten.
Wrong working
$(cos x)^{'} = sin x$ .
Right working
$(cos x)^{'} = - sin x$ .
Do this: Remember the "co" functions get a minus sign.
Usually costs: 1 mark.

Exam tips

In first-principle questions write the difference quotient on one line and state "as h tends to 0" at the end.
Label u, v, u', v' in a small table before applying the product or quotient rule.
After a quotient rule, simplify with identities such as $sin^{2} x + cos^{2} x = 1$ ; many answers collapse.
Write the tangent equation in the form $y - y_{1} = m (x - x_{1})$ and state the point first.
The chain rule is only formative in 2026-27: expand or use identities instead.

Practice questions on Derivatives: First Principle and Rules of Differentiation

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
If $f (x) = x^{5} - 3 x^{2} + 7$ , then $f^{'} (1)$ equals:
(a) $- 1$
(b) 1
(c) 5
(d) 8
Hint
$f^{'} (x) = 5 x^{4} - 6 x$ .
Answer
(a). $5 - 6 = - 1$ .
Q2 · 1 mark · MCQHarder board level
If $f (x) = x tan x$ , then $f^{'} (\frac{π}{4})$ equals:
(a) $1 + \frac{π}{2}$
(b) $1 + \frac{π}{4}$
(c) $1 + π$
(d) $2 + \frac{π}{2}$
Hint
$f^{'} (x) = tan x + x sec^{2} x$ .
Answer
(a). $tan \frac{π}{4} = 1$ , $sec^{2} \frac{π}{4} = 2$ , so $1 + \frac{π}{4} \cdot 2 = 1 + \frac{π}{2}$ .
Q3 · 3 marks · Short answerHarder board level
Differentiate $\frac{sin x + cos x}{sin x - cos x}$ and simplify.
Hint
The numerator of the quotient rule simplifies using $sin^{2} x + cos^{2} x = 1$ .
Answer
Numerator: $(cos x - sin x) (sin x - cos x) - (sin x + cos x) (cos x + sin x) = - (sin x - cos x)^{2} - (sin x + cos x)^{2} = - 2$ . So the derivative is $\frac{- 2}{( sin x - cos x ) ^{2}}$ .
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): $\frac{d}{d x} (sin x)$ at $x = 0$ is 1.
Reason (R): $h \to 0 lim \frac{sin h}{h} = 1$ .
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Use first principle at $x = 0$ .
Answer
(a). $f^{'} (0) = lim \frac{s i n h - s i n 0}{h} = lim \frac{s i n h}{h} = 1$ .
Q5 · 4 marks · Short answerHOTS / competency
Find the points on the curve $y = x^{3} - 3 x + 2$ at which the tangent is parallel to the line $y = 9 x + 5$ , and write the equation of the tangent at the point with positive $x$ .
Hint
Parallel means equal slope: $f^{'} (x) = 9$ .
Answer
$f^{'} (x) = 3 x^{2} - 3 = 9$ , so $x^{2} = 4$ , $x = \pm 2$ . $y (2) = 8 - 6 + 2 = 4$ and $y (- 2) = - 8 + 6 + 2 = 0$ . Points $(2, 4)$ and $(- 2, 0)$ . Tangent at $(2, 4)$ : $y - 4 = 9 (x - 2)$ , i.e. $y = 9 x - 14$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.What is the definition of f'(x)?Show answer
lim (f(x + h) − f(x))/h as h tends to 0.
2.What is the derivative of x^n?Show answer
n x^(n−1).
3.What is the product rule?Show answer
(uv)' = u'v + uv'.
4.What is the derivative of cos x?Show answer
−sin x.
5.What does f'(a) represent for the graph?Show answer
The slope of the tangent at x = a.

Derivatives: First Principle and Rules of Differentiation: frequently asked questions

When must I use first principle?

Only when the question says "from first principle" or "using the definition". Otherwise use the rules.

Is the chain rule in the 2026-27 syllabus?

The chapter page notes that the chain rule for composite functions is assessed only formatively, so annual-exam questions stay with the power, sum, product and quotient rules.

What is the difference between average and instantaneous rate?

Average rate is a ratio over an interval; instantaneous rate is its limit as the interval shrinks, which is the derivative.

Can I use the derivative to find velocity?

Yes. If s(t) is position, v = ds/dt and acceleration a = dv/dt.