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State Board · Class 11 · Mathematics Part 2

Lesson 9: Differentiation

5 concepts · 40 practice questions with answers · Free to read

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Meaning of the derivative

From daily life

A speedometer shows how fast the distance changes at one instant. The derivative likewise tells how steep a curve is at a single point. Building on classes 9–10 : Since classes 9–10 we know the slope of a line; here we find the slope of a curve.

Simple definition

The derivative of f at x is f′(x) = lim [f(x + h) − f(x)]/h as h → 0. It gives the slope of the tangent at that point.

Rule 1: Derivative from first principles — Write f(x + h) − f(x), divide by h, simplify, then let h → 0. The derivative of a constant is 0.

PQQ → Psecant → tangentslope = lim [f(x+h) − f(x)]/h

In the figure : As the secant closes in it becomes the tangent.

How is the formula obtained? : f(x) = x²: [(x + h)² − x²]/h = (2xh + h²)/h = 2x + h; letting h → 0 gives f′(x) = 2x.

  • Easy : f(x) = 3x + 2: [3(x + h) + 2 − 3x − 2]/h = 3; f′(x) = 3.
  • Medium : f(x) = x³: [(x + h)³ − x³]/h = 3x² + 3xh + h²; f′(x) = 3x².
  • Exam-type : f(x) = 1/x: (1/(x + h) − 1/x)/h = −1/(x(x + h)); f′(x) = −1/x².
Remember

Derivative = rate of change = slope of the tangent.

Common mistake

Putting h = 0 too early gives 0/0; simplify and cancel h first.

Shortcut

Differentiate first and put the value of x afterwards. Memory clue : "Derivative first, value next".

CET trap

f′(5) means putting x = 5 in f′(x), not differentiating the number f(5) (which would give 0).

CET-style example : If f(x) = x², then f′(5) = ? (A) 10 (B) 25 (C) 5 (D) 2

Answer : (A). f′(x) = 2x; 2 × 5 = 10. Time-saving method : Put the value straight into 2x.

Spot the difference

TermMeaning
Slope of secant[f(x + h) − f(x)]/h
Slope of tangentf′(x) (h → 0)

Practice questions

Easy
  1. 1What is the derivative of the constant 5?
  2. 2What is the derivative of f(x) = 7x?
  3. 3Find the derivative of f(x) = x² at x = 3.
Medium
  1. 4Find the derivative of f(x) = x² + 3x at x = 2.
  2. 5Find the derivative of f(x) = 2x² at x = 3.
  3. 6Find the derivative of f(x) = x² − 4 at x = 5.
Challenging
  1. 7Find the derivative of f(x) = x³ at x = 2.
  2. 8Find the derivative of f(x) = 1/x at x = 2.
Show answers (try them yourself first)
  1. 10.
  2. 27.
  3. 36.
  4. 42x + 3 ; 2 · 2 + 3 = 7.
  5. 54x ; 12.
  6. 62x ; 10.
  7. 73x² ; 12.
  8. 8−1/x² ; −1/4.

Differentiability and continuity

From daily life

At a sharp turn on a road the direction changes suddenly and there is no single tangent. Such a corner on a graph means the function is not differentiable. Building on classes 9–10 : Since classes 9–10 we have seen the V-shaped graph of ∣x∣; here we test its corner.

Simple definition

f is differentiable at a if the left derivative (LHD) and the right derivative (RHD) exist and are equal. Differentiable ⇒ continuous; but continuous does not imply differentiable.

Rule 1: The test — LHD = lim [f(a + h) − f(a)]/h as h → 0⁻, RHD = the same quotient as h → 0⁺; differentiable only if both are equal.

slope −1slope +1left slope ≠ right slope at 0continuous, not differentiable

In the figure : The corner of ∣x∣ at 0: LHD = −1, RHD = 1.

How is the formula obtained? : For x < 0, ∣x∣ = −x with slope −1; for x > 0, ∣x∣ = x with slope 1. They differ, so there is no tangent at 0.

  • Easy : f(x) = ∣x∣ at 0: LHD = −1, RHD = 1; not differentiable, but continuous.
  • Medium : f(x) = ∣x − 2∣ at 2: LHD = −1, RHD = 1; not differentiable.
  • Exam-type : f(x) = x² at 3: LHD = RHD = 6; differentiable, f′(3) = 6.
Remember

Differentiable always means continuous; not the other way round.

Common mistake

Believing "continuous means differentiable"; remember ∣x∣ as the counter-example.

Shortcut

Wherever you see a corner or a break compute the LHD and RHD. Memory clue : "A corner means not differentiable".

CET trap

A discontinuous function is never differentiable.

CET-style example : The derivative of f(x) = ∣x∣ at x = 0? (A) does not exist (B) 0 (C) 1 (D) −1

Answer : (A). LHD = −1, RHD = 1. Time-saving method : The two differ: it does not exist.

Spot the difference

PropertyMeaning
Continuousgraph is unbroken
Differentiablegraph is unbroken and has no corner

Practice questions

Easy
  1. 1Is ∣x∣ continuous at 0?
  2. 2Is ∣x∣ differentiable at 0?
  3. 3Is a differentiable function continuous?
Medium
  1. 4Find the LHD and RHD of ∣x + 1∣ at x = −1.
  2. 5Find the LHD and RHD of 3∣x∣ at x = 0.
  3. 6Find the RHD of x² at x = 2.
Challenging
  1. 7f(x) = x (x < 1), 2x − 1 (x ≥ 1): is it continuous at x = 1? Find the LHD and RHD.
  2. 8What can be said about the differentiability of this function at x = 1?
Show answers (try them yourself first)
  1. 1होय.
  2. 2नाही.
  3. 3होय.
  4. 4LHD −1, RHD 1.
  5. 5LHD −3, RHD 3.
  6. 64.
  7. 7संतत ; LHD 1, RHD 2.
  8. 8अवकलनीय नाही (LHD ≠ RHD).

Rules of differentiation

From daily life

Taking the derivative of a long sum, product or quotient from first principles each time is slow; so ready rules are used. Building on classes 9–10 : Since classes 9–10 we know powers and algebraic expressions; here are the rules to differentiate them.

Simple definition

Power rule: (xⁿ)′ = n xⁿ⁻¹. Sum: (u ± v)′ = u′ ± v′. Product: (uv)′ = u′v + uv′. Quotient: (u/v)′ = (u′v − uv′)/v².

Rule 1: Using the rules — Split into terms and use the power rule; for a product take the first × derivative of the second + the second × derivative of the first. In a quotient square the denominator.

u vv duu dvuvd(uv) =u dv + v du(uv)′ = u v′ + v u′

In the figure : The area of a product and its two thin growth strips.

How is the formula obtained? : When the length u and width v of a rectangle grow, the area grows by two strips, u′v and uv′; hence (uv)′ = u′v + uv′.

  • Easy : (3x⁴ − 5x² + 7)′ = 12x³ − 10x.
  • Medium : ((x² + 1)(x³ − 2))′ = 2x(x³ − 2) + (x² + 1)3x² = 5x⁴ + 3x² − 4x; at x = 1 it is 4.
  • Exam-type : ((2x + 3)/(x − 1))′ = (2(x − 1) − (2x + 3))/(x − 1)² = −5/(x − 1)².
Remember

The derivative of a constant is 0; (c·u)′ = c·u′.

Common mistake

Writing (uv)′ = u′v′; use the product rule.

Shortcut

If you first expand a product into a polynomial the power rule alone is enough. Memory clue : "First-second, second-first".

CET trap

Do not swap the order in the quotient numerator: u′v − uv′.

CET-style example : The derivative of x⁵ at x = 2? (A) 80 (B) 32 (C) 10 (D) 40

Answer : (A). 5x⁴ = 5 · 16 = 80. Time-saving method : Bring the power down, reduce it by one.

Spot the difference

RuleFormula
Productu′v + uv′
Quotient(u′v − uv′)/v²

Practice questions

Easy
  1. 1Find the derivative of 4x³ at x = 1.
  2. 2Find the derivative of x² + 5x at x = 0.
  3. 3What is the derivative of the constant 9?
Medium
  1. 4Find the derivative of (x + 1)(x − 2) at x = 3.
  2. 5Find the derivative of x/(x + 1) at x = 1.
  3. 6Find the derivative of (2x + 1)² at x = 1.
Challenging
  1. 7Find the derivative of (x² + 1)/(x² − 1) at x = 2.
  2. 8Find the derivative of x²(x + 1) at x = 1.
Show answers (try them yourself first)
  1. 112.
  2. 25.
  3. 30.
  4. 42x − 1 ; 5.
  5. 51/(x + 1)² ; 1/4.
  6. 64(2x + 1) ; 12.
  7. 7−4x/(x² − 1)² ; −8/9.
  8. 8x³ + x² चा अवकलज 3x² + 2x ; 3 + 2 = 5.

Derivatives of trigonometric functions

From daily life

The height of a point on a wheel rises and falls like a wave; the derivatives of sin and cos tell how fast that height changes at each moment. Building on classes 9–10 : Since classes 9–10 we know sin, cos and tan; here are their derivatives.

Simple definition

(sin x)′ = cos x, (cos x)′ = −sin x, (tan x)′ = sec² x, (cot x)′ = −cosec² x, (sec x)′ = sec x tan x, (cosec x)′ = −cosec x cot x. (x in radians).

Rule 1: Using them — Use them with the product and quotient rules. Always take x in radians.

slope 1slope 0slope −1(sin x)′ = cos x

In the figure : The slope of sin is cos.

How is the formula obtained? : The graph of sin has its steepest slope (1) at 0, where cos is 1; at the top of sin the slope is 0 and cos is 0 there; hence (sin x)′ = cos x.

  • Easy : (sin x)′ = cos x; at x = 0 it is 1.
  • Medium : (x sin x)′ = sin x + x cos x; at x = π/2 it is 1 + 0 = 1.
  • Exam-type : (sin x/x)′ = (x cos x − sin x)/x²; at x = π it is (−π − 0)/π² = −1/π.
Remember

Derivatives of the "co" functions are negative.

Common mistake

Writing (cos x)′ = sin x; do not forget the minus sign.

Shortcut

The cycle sin → cos → −sin → −cos → sin. Memory clue : "s, c, −s, −c: four steps, then repeat".

CET trap

These formulas hold only when x is in radians; in degrees they change.

CET-style example : The derivative of (sin x + cos x) at x = π/4? (A) 0 (B) 1 (C) √2 (D) −1

Answer : (A). cos(π/4) − sin(π/4) = 0. Time-saving method : Both are 1/√2; the difference is 0.

Spot the difference

FunctionDerivative
sin xcos x
cos x−sin x
tan xsec² x

Practice questions

Easy
  1. 1Find the value of (sin x)′ at x = π/3.
  2. 2Find the value of (cos x)′ at x = π/6.
  3. 3Find the value of (cos x)′ at x = 0.
Medium
  1. 4Find the derivative of 2 sin x + 3 cos x at x = π/2.
  2. 5Find the derivative of tan x at x = π/4.
  3. 6Find the derivative of x cos x at x = 0.
Challenging
  1. 7Find the derivative of sin x cos x at x = π/4.
  2. 8Find the derivative of tan x at x = π/3.
Show answers (try them yourself first)
  1. 11/2.
  2. 2−1/2.
  3. 30.
  4. 42 cos x − 3 sin x ; −3.
  5. 5sec²(π/4) = 2.
  6. 6cos x − x sin x ; 1.
  7. 7cos 2x ; 0.
  8. 8sec²(π/3) = 4.

Derivatives of exponential and logarithmic functions

From daily life

In compound interest the larger the amount, the faster it grows. eˣ is the function whose rate of growth equals its own value. Building on classes 9–10 : Since classes 9–10 we know exponents and logarithms; here are their derivatives.

Simple definition

(eˣ)′ = eˣ, (aˣ)′ = aˣ log a, (log x)′ = 1/x (x > 0).

Rule 1: Using them — Use them with the product and quotient rules. Here log means the natural logarithm (base e).

(0, 1)height e, slope e(eˣ)′ = eˣ

In the figure : The slope of eˣ equals its height.

How is the formula obtained? : At any point of the graph of eˣ the slope and the height are equal; hence (eˣ)′ = eˣ. Since aˣ = e^(x log a), (aˣ)′ = aˣ log a.

  • Easy : (3eˣ)′ = 3eˣ; at x = 0 it is 3.
  • Medium : (x eˣ)′ = eˣ + x eˣ = (1 + x)eˣ; at x = 1 it is 2e ≈ 5.4366.
  • Exam-type : (x² log x)′ = 2x log x + x; at x = e it is 2e + e = 3e ≈ 8.155.
Remember

(eˣ)′ = eˣ: this is the function that is its own derivative.

Common mistake

Writing (aˣ)′ = x aˣ⁻¹; in an exponential the variable is in the power, the base is constant.

Shortcut

Taking a = e in (aˣ)′ gives log e = 1 and hence eˣ. Memory clue : "Exponential stays itself, only multiply by log a".

CET trap

(log x)′ = 1/x only for x > 0.

CET-style example : The derivative of x eˣ at x = 1? (A) 2e (B) e (C) 1 + e (D) 2

Answer : (A). (1 + x)eˣ = 2e. Time-saving method : Product rule.

Spot the difference

FunctionDerivative
eˣeˣ
aˣaˣ log a
log x1/x

Practice questions

Easy
  1. 1Find the derivative of 5eˣ at x = 0.
  2. 2Find the derivative of log x at x = 2.
  3. 3Find the derivative of 2ˣ at x = 3.
Medium
  1. 4Find the derivative of eˣ log x at x = 1.
  2. 5Find the derivative of 3eˣ − 2 log x at x = 1.
  3. 6Find the derivative of eˣ sin x at x = 0.
Challenging
  1. 7Find the derivative of (log x)/x at x = e.
  2. 8Find the derivative of x log x − x at x = 3.
Show answers (try them yourself first)
  1. 15.
  2. 21/2.
  3. 38 log 2 ≈ 5.545.
  4. 4eˣ log x + eˣ/x ; e.
  5. 53e − 2 ≈ 6.1548.
  6. 6eˣ(sin x + cos x) ; 1.
  7. 7(1 − log x)/x² ; 0.
  8. 8log x ; log 3.

Lesson 9: Differentiation — quick revision

TopicFormula
Definitionf′(x) = lim [f(x + h) − f(x)]/h
Power(xⁿ)′ = n xⁿ⁻¹
Product(uv)′ = u′v + uv′
Quotient(u/v)′ = (u′v − uv′)/v²
sin, coscos x, −sin x
Exponential, logeˣ, aˣ log a, 1/x

🎯 Most important points for the exam

  • Differentiable means LHD = RHD.
  • Differentiable implies continuous.
  • A product gives two terms.
  • A quotient has the square of the denominator.
  • In trigonometry x is in radians.
Study mantra

"Derivative first, value next".