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Mathematics

Differentiation

📘 Calculus · Chapter MN12 🎯 NDA Level : High Priority

Differentiation is the process of finding the rate at which a function changes. It is the operational core of calculus — every question on tangents, rates, maxima, and minima builds on derivatives. For NDA, this chapter is high-yield: questions are direct, formula-driven, and reward systematic application of the differentiation rules.

📌 What to expect in NDA (based on 2022–2024 papers):
(1) Finding dy/dx for standard functions (xⁿ, sin x, e^x, ln x) directly;
(2) Applying product rule to products like x² sin x or e^x ln x;
(3) Quotient rule for fractions like sin x / x or (x²+1)/(x−1);
(4) Chain rule for composite functions like sin(x²), e^{sin x}, ln(cos x);
(5) Implicit differentiation when y is not isolated (x² + y² = r² type);
(6) Parametric differentiation when x = f(t) and y = g(t);
(7) Logarithmic differentiation for x^x or y = (sin x)^{cos x};
(8) Second derivative d²y/dx² — finding it from dy/dx.

Topics at a Glance

① First Principles

Limit definition, slope of tangent

② Standard Derivatives

xⁿ, sin x, cos x, e^x, ln x, a^x

③ Differentiation Rules

Sum, product, quotient, chain

④ Implicit Differentiation

Differentiate both sides w.r.t. x

⑤ Parametric & Log Diff.

dy/dx = (dy/dt)/(dx/dt); log both sides

⑥ Higher-Order Derivatives

d²y/dx², finding n^th derivative

1. Derivative by First Principles

1.1

The Fundamental Definition of a Derivative

The derivative is the slope of the tangent — found as the limit of a difference quotient

The derivative of f(x) at x = a is the instantaneous rate of change — geometrically, the slope of the tangent line to the curve at that point. It is defined as a limit.

⚡ Derivative — First Principles Definition

f′(x) = lim[h→0] [f(x + h) − f(x)] / h (standard form) Also written as: dy/dx = lim[δx→0] δy / δx = lim [f(x+δx) − f(x)] / δx Equivalent notations for the derivative: f′(x) = dy/dx = y′ = D[f(x)] = (d/dx) f(x) Geometric meaning: f′(a) = slope of tangent to y = f(x) at the point (a, f(a)) Positive f′(a) → function is increasing at x = a Negative f′(a) → function is decreasing at x = a f′(a) = 0 → possible maximum, minimum, or inflection point

For NDA, first principles is mostly tested for simple functions (x², x³, sin x) to derive the formula, not to evaluate complex derivatives. Know the definition and how to apply it for 2–3 step problems.

Fig 1: As h → 0, Q approaches P and the secant PQ becomes the tangent at P. The slope of the tangent = f′(x) = dy/dx.

Worked Example — Derivative of x² from First Principles

Find f′(x) for f(x) = x² using the definition.

f′(x) = lim_h→0 [(x+h)² − x²] / h

= lim_h→0 [x² + 2xh + h² − x²] / h = lim_h→0 (2x + h) = 2x.

Worked Example — Derivative of sin x from First Principles

Find d(sin x)/dx using first principles.

f′(x) = lim_h→0 [sin(x+h) − sin x] / h

Using sin(x+h) = sin x cos h + cos x sin h:

= lim_h→0 [sin x cos h + cos x sin h − sin x] / h

= lim_h→0 [sin x (cos h − 1)/h + cos x ⋅ sin h/h]

= sin x ⋅ 0 + cos x ⋅ 1 = cos x.

Used: lim (cos h − 1)/h = 0 and lim sin h/h = 1 as h→0.

2. Standard Derivatives — The Core Table

2.1

Derivatives of All Standard Functions

Memorise this table completely — every differentiation rule builds on these

⚡ Complete Standard Derivatives Table

ALGEBRAIC: d/dx (c) = 0 (c is a constant) d/dx (x) = 1 d/dx (x²) = 2x d/dx (x³) = 3x² d/dx (xⁿ) = n x^(n-1) (Power Rule, valid for all real n) d/dx (1/x) = -1/x² [= d/dx (x^-1) = -x^-2] d/dx (√x) = 1/(2√x) [= d/dx (x^(1/2))] TRIGONOMETRIC: d/dx (sin x) = cos x d/dx (cos x) = -sin x d/dx (tan x) = sec² x d/dx (cot x) = -cosec² x d/dx (sec x) = sec x tan x d/dx (cosec x) = -cosec x cot x INVERSE TRIGONOMETRIC: d/dx (sin⁻¹ x) = 1/√(1-x²) d/dx (cos⁻¹ x) = -1/√(1-x²) d/dx (tan⁻¹ x) = 1/(1+x²) EXPONENTIAL & LOGARITHMIC: d/dx (eˣ) = eˣ (only function equal to its own derivative!) d/dx (aˣ) = aˣ ln a (a > 0, a ≠ 1) d/dx (ln x) = 1/x (x > 0) d/dx (logₔ x) = 1/(x ln a)

Power rule trick: bring the exponent down as a coefficient, reduce the exponent by 1. Works for ALL rational powers. Example: d/dx(x^(-3)) = -3x^(-4).

Signs to Remember

d/dx sin x = + cos x
d/dx cos x = − sin x (negative!)
d/dx tan x = + sec² x
d/dx cot x = − cosec² x (negative!)
d/dx sec x = + sec x tan x
d/dx cosec x = − cosec x cot x (negative!)

Power Rule Quick Examples

d/dx (x⁵) = 5x⁴
d/dx (x^(1/3)) = (1/3)x^(-2/3)
d/dx (x^(-2)) = -2x^(-3) = -2/x³
d/dx (√x) = 1/(2√x)
d/dx (1/x²) = d/dx(x^-2) = -2/x³

📝 TOPIC-WISE PYQ

Standard Derivatives — NDA-Pattern Questions

Q1. Differentiate f(x) = x³ − 4x² + 7x − 2 with respect to x.

(a) 3x² − 8x + 7 (b) 3x² − 4x + 7 (c) 3x² − 8x − 7 (d) x² − 8x + 7

Answer: (a) 3x² − 8x + 7
d/dx(x³) = 3x², d/dx(-4x²) = -8x, d/dx(7x) = 7, d/dx(-2) = 0.
f′(x) = 3x² − 8x + 7.

Q2. If y = sin x + cos x, then dy/dx is:

(a) cos x − sin x (b) cos x + sin x (c) −sin x + cos x (d) −cos x − sin x

Answer: (a) cos x − sin x
d/dx(sin x) = cos x, d/dx(cos x) = −sin x.
dy/dx = cos x − sin x.

Q3. The derivative of e^x + ln x + √x is:

(a) e^x + 1/x + 1/√x (b) e^x + 1/x + 1/(2√x) (c) e^x − 1/x + 1/(2√x) (d) xe^x + 1/x

Answer: (b) e^x + 1/x + 1/(2√x)
d/dx(e^x) = e^x, d/dx(ln x) = 1/x, d/dx(x^1/2) = (1/2)x^−1/2 = 1/(2√x).
Result: e^x + 1/x + 1/(2√x).

3. Rules of Differentiation

3.1

Sum, Product, Quotient & Chain Rules

Four rules that let you differentiate any combination of standard functions

➕ Sum / Difference Rule

d/dx [f ± g] = f′ ± g′

d/dx[x² + sin x] = 2x + cos x

✖ Product Rule (uv)

d/dx[uv] = u′v + uv′

d/dx[x² sin x] = 2x sin x + x² cos x

÷ Quotient Rule (u/v)

d/dx[u/v] = (u′v − uv′) / v²

d/dx[sin x / x] = (x cos x − sin x) / x²

🔗 Chain Rule (composite)

d/dx[f(g(x))] = f′(g(x)) ⋅ g′(x)

d/dx[sin(x²)] = cos(x²) ⋅ 2x

⚡ Rules Expanded — Careful Form

PRODUCT RULE: If y = u ⋅ v where u, v are functions of x: dy/dx = u (dv/dx) + v (du/dx) Memory: "first times derivative of second, plus second times derivative of first" QUOTIENT RULE: If y = u/v: dy/dx = [v(du/dx) − u(dv/dx)] / v² Memory: "bottom times d(top) minus top times d(bottom), all over bottom squared" Note: denominator is v² (NOT v), numerator sign is MINUS (NOT plus) CHAIN RULE — Step by Step: Step 1: Identify outer function f and inner function g(x) Step 2: Differentiate outer f, keeping inner unchanged Step 3: Multiply by derivative of inner g′(x) Example: y = (3x²+1)&sup5; Outer: u&sup5; where u = 3x²+1 dy/dx = 5(3x²+1)⁴ ⋅ 6x = 30x(3x²+1)⁴

Chain rule is by far the most tested rule in NDA. Any function of a function — sin(x²), e^3x, ln(cos x), (x²+1)&sup5; — requires the chain rule. Always write outer derivative first, then multiply by inner derivative.

Worked Example — Product Rule

Find dy/dx if y = x² ln x.

u = x², u′ = 2x. v = ln x, v′ = 1/x.

dy/dx = u′v + uv′ = 2x ⋅ ln x + x² ⋅ (1/x) = 2x ln x + x = x(2 ln x + 1).

Worked Example — Quotient Rule

Differentiate y = (x²+1)/(x−1).

u = x²+1, u′ = 2x. v = x−1, v′ = 1.

dy/dx = [(x−1)(2x) − (x²+1)(1)] / (x−1)²

= [2x² − 2x − x² − 1] / (x−1)² = (x² − 2x − 1) / (x−1)².

Worked Example — Chain Rule (three layers)

Find d/dx [e^sin(x²)].

Outermost: e^u, u = sin(x²). d/dx[e^u] = e^u ⋅ du/dx.

du/dx = d/dx[sin(x²)] = cos(x²) ⋅ 2x. (chain rule again for the inner part)

Final: e^sin(x²) ⋅ 2x cos(x²).

📝 TOPIC-WISE PYQ

Differentiation Rules — NDA-Pattern Questions

Q4. Find d/dx[x² e^x].

(a) 2x e^x (b) x e^x(x+2) (c) xe^x(x+2) (d) e^x(x²+2x)

Answer: (d) e^x(x²+2x)
Product rule: u=x², u′=2x; v=e^x, v′=e^x.
dy/dx = 2x⋅e^x + x²⋅e^x = e^x(2x+x²) = e^x(x²+2x).

Q5. If y = sin(3x²+2), find dy/dx.

(a) cos(3x²+2) (b) 6x cos(3x²+2) (c) −6x cos(3x²+2) (d) 6x sin(3x²+2)

Answer: (b) 6x cos(3x²+2)
Chain rule: outer = sin(u), inner u = 3x²+2.
dy/dx = cos(3x²+2) ⋅ d/dx(3x²+2) = cos(3x²+2) ⋅ 6x = 6x cos(3x²+2).

Q6. Differentiate y = tan x / (1 + sec x).

(a) 1/(1+sec x) (b) sec x/(1+sec x) (c) 1/(sec x+1)² (d) cosec x

Answer: (a) 1/(1+sec x)
u=tan x, u′=sec²x; v=1+sec x, v′=sec x tan x.
dy/dx = [sec²x(1+sec x) − tan x ⋅ sec x tan x] / (1+sec x)²
= [sec²x + sec³x − sec x tan²x] / (1+sec x)².
Using tan²x = sec²x−1: numerator = sec²x + sec x(sec²x−(sec²x−1)) = sec²x + sec x = sec x(sec x+1).
dy/dx = sec x(1+sec x)/(1+sec x)² = sec x/(1+sec x). Option (b).

Q7. d/dx [ln(sin x)] = ?

(a) cos x / sin x (b) 1/sin x (c) cot x (d) −cot x

Answer: (c) cot x
Chain rule: outer = ln(u), inner u = sin x.
d/dx = (1/sin x) ⋅ cos x = cos x/sin x = cot x.

🔥 TRICKY QUESTIONS

Rules — Nested Chain & Mixed Rules

🤯 T1. Find d/dx[ln(sin(e^x))].

Three-layer chain rule: outer = ln(u), middle = sin(v), inner = e^x.
d/dx = (1/sin(e^x)) ⋅ cos(e^x) ⋅ e^x
= e^x cos(e^x) / sin(e^x) = e^x cot(e^x).
Strategy: Work layer by layer from outside in. Each time, multiply by the derivative of the next inner layer.

🤯 T2. If y = x sin x cos x, find dy/dx.

First rewrite: y = x ⋅ (sin x cos x) = (x/2) sin(2x) using sin(2x) = 2 sin x cos x.
dy/dx = (1/2)[sin(2x) + x ⋅ 2cos(2x)] = (1/2)[sin 2x + 2x cos 2x]
= (sin 2x)/2 + x cos 2x.
Alternatively by product rule twice: y = x ⋅ u where u = sin x cos x. u′ = cos²x − sin²x = cos 2x.
dy/dx = sin x cos x + x cos 2x = (sin 2x)/2 + x cos 2x. Same answer ✓.

4. Implicit Differentiation

4.1

Differentiating When y Cannot Be Isolated

Differentiate every term w.r.t. x; treat y as a function of x using the chain rule

When a curve is given in implicit form F(x, y) = 0, we cannot write y explicitly as f(x). Instead, differentiate both sides of the equation with respect to x, applying the chain rule to every term involving y.

⚡ Implicit Differentiation — Method

Key rule: d/dx[yⁿ] = n y^(n-1) ⋅ dy/dx (chain rule applied to y) General procedure: Step 1: Differentiate every term of F(x, y) = 0 with respect to x. Step 2: For any term with y: • d/dx (y) = dy/dx • d/dx (y²) = 2y ⋅ dy/dx • d/dx (y³) = 3y² ⋅ dy/dx • d/dx (sin y) = cos y ⋅ dy/dx • d/dx (xy) = x ⋅ dy/dx + y [product rule + chain] Step 3: Collect all terms with dy/dx on one side. Step 4: Factor out dy/dx and solve.

The only difference from explicit differentiation is that y-terms always carry a factor of dy/dx from the chain rule. Collect and solve algebraically at the end.

Worked Example — Circle x² + y² = r²

Find dy/dx for x² + y² = r².

Differentiate both sides w.r.t. x:

2x + 2y ⋅ dy/dx = 0.

2y ⋅ dy/dx = −2x.

dy/dx = −x/y. (Geometric: slope of tangent to circle = −x/y.)

Worked Example — Mixed Implicit

Find dy/dx if x² + xy + y² = 7.

Differentiate: 2x + (y + x⋅dy/dx) + 2y⋅dy/dx = 0.

2x + y + dy/dx(x + 2y) = 0.

dy/dx = −(2x + y) / (x + 2y).

📝 TOPIC-WISE PYQ

Implicit Differentiation — NDA-Pattern Questions

Q8. If x² + y² = 25, find dy/dx at (3, 4).

(a) −3/4 (b) 3/4 (c) −4/3 (d) 4/3

Answer: (a) −3/4
From implicit differentiation: dy/dx = −x/y.
At (3, 4): dy/dx = −3/4 = −3/4.

Q9. Find dy/dx if sin(x+y) = y.

(a) cos(x+y) / [1−cos(x+y)] (b) cos(x+y) (c) 1/[1−cos(x+y)] (d) −cos(x+y)

Answer: (a) cos(x+y) / [1−cos(x+y)]
Differentiate: cos(x+y) ⋅ (1 + dy/dx) = dy/dx.
cos(x+y) + cos(x+y)⋅dy/dx = dy/dx.
cos(x+y) = dy/dx − cos(x+y)⋅dy/dx = dy/dx[1−cos(x+y)].
dy/dx = cos(x+y)/[1−cos(x+y)].

🔥 TRICKY QUESTIONS

Implicit Differentiation — Classic NDA Traps

🤯 T3. If y = √(x + √(x + √x ...)) (infinite nested radical), find dy/dx.

Since the radical repeats infinitely: y = √(x + y). (The pattern repeats, so y appears inside itself.)
Squaring: y² = x + y.
Implicit differentiation: 2y⋅dy/dx = 1 + dy/dx.
dy/dx(2y − 1) = 1.
dy/dx = 1/(2y − 1).
Key insight: Recognise the infinite nested pattern and express y in terms of itself. This is a classic trick that appears in NDA and competitive exams.

5. Parametric & Logarithmic Differentiation

5.1

Parametric Differentiation — dy/dx via a Parameter t

When x = f(t) and y = g(t), divide dy/dt by dx/dt

⚡ Parametric Differentiation Formula

If x = f(t) and y = g(t), then: dy/dx = (dy/dt) / (dx/dt) provided dx/dt ≠ 0 Steps: 1. Find dy/dt (differentiate y w.r.t. t) 2. Find dx/dt (differentiate x w.r.t. t) 3. Divide: dy/dx = (dy/dt) ÷ (dx/dt) Standard parametric pairs: Circle: x = r cos t, y = r sin t → dy/dx = −sin t / cos t = −tan t Parabola: x = at², y = 2at → dy/dx = 2a / 2at = 1/t Ellipse: x = a cos t, y = b sin t → dy/dx = −b cos t / (a sin t)

Parametric is tested in NDA typically with the cycloid or circle. The formula dy/dx = (dy/dt)/(dx/dt) is a direct division of two ordinary derivatives — simple to apply once both dt derivatives are found.

Worked Example — Parametric

x = a(t − sin t), y = a(1 − cos t). Find dy/dx.

dx/dt = a(1 − cos t). dy/dt = a sin t.

dy/dx = a sin t / a(1 − cos t) = sin t / (1 − cos t).

Using half-angle: sin t = 2 sin(t/2) cos(t/2), 1−cos t = 2 sin²(t/2).

dy/dx = 2 sin(t/2) cos(t/2) / 2sin²(t/2) = cot(t/2).

5.2

Logarithmic Differentiation — For x^x and Exponential-Form Functions

Take ln of both sides, differentiate, then multiply by y

Logarithmic differentiation is used when the function has a variable both in the base and the exponent, like y = x^x or y = (sin x)^{cos x}. Taking ln first converts the exponent into a product.

⚡ Logarithmic Differentiation — Method

For y = [f(x)]^[g(x)]: Step 1: Take ln of both sides: ln y = g(x) ⋅ ln[f(x)] Step 2: Differentiate both sides w.r.t. x: (1/y) ⋅ dy/dx = g′(x) ⋅ ln[f(x)] + g(x) ⋅ f′(x)/f(x) Step 3: Multiply both sides by y: dy/dx = y ⋅ [g′(x) ⋅ ln f(x) + g(x) ⋅ f′(x)/f(x)] Special case — y = xˣ: ln y = x ln x (1/y) dy/dx = ln x + x ⋅ (1/x) = ln x + 1 dy/dx = xˣ(1 + ln x) Also useful for products/quotients of many functions: y = [u(x) ⋅ v(x)] / w(x) ln y = ln u + ln v − ln w (1/y) dy/dx = u′/u + v′/v − w′/w

Logarithmic differentiation is tested in NDA for x^x-type derivatives and for simplifying products of many functions. Always take ln, differentiate, then multiply back by y.

Worked Example — y = (sin x)^{tan x}

Find dy/dx for y = (sin x)^{tan x}.

ln y = tan x ⋅ ln(sin x).

(1/y) dy/dx = sec²x ⋅ ln(sin x) + tan x ⋅ (cos x/sin x).

= sec²x ln(sin x) + tan x ⋅ cot x = sec²x ln(sin x) + 1.

dy/dx = y[sec²x ln(sin x) + 1] = (sin x)^{tan x}[1 + sec²x ln(sin x)].

📝 TOPIC-WISE PYQ

Parametric & Logarithmic Differentiation — NDA-Pattern Questions

Q10. If x = a cos t, y = b sin t, find dy/dx.

(a) −b cos t / (a sin t) (b) b cos t / (a sin t) (c) −(b/a) cot t (d) (b/a) cot t

Answer: (a) −b cos t / (a sin t)
dx/dt = −a sin t, dy/dt = b cos t.
dy/dx = b cos t / (−a sin t) = −(b/a) cot t. (Options (a) and (c) are equivalent.)

Q11. If y = x^x, then dy/dx is:

(a) x^x−1 (b) x^x ln x (c) x^x(1 + ln x) (d) x^x(ln x − 1)

Answer: (c) x^x(1 + ln x)
ln y = x ln x. (1/y) dy/dx = ln x + x(1/x) = ln x + 1.
dy/dx = y(1 + ln x) = x^x(1 + ln x).
Common trap: (a) x^x−1 uses the power rule — ONLY valid when the exponent is a constant, not variable.

Q12. If x = t² and y = t³, find dy/dx.

(a) t/2 (b) 3t/2 (c) 2/3t (d) 3t²/2t

Answer: (b) 3t/2
dx/dt = 2t, dy/dt = 3t².
dy/dx = 3t² / 2t = 3t/2.

🔥 TRICKY QUESTIONS

Parametric & Log Diff. — Harder NDA Problems

🤯 T4. Differentiate y = x^{sin x} + (sin x)^x.

Let u = x^{sin x} and v = (sin x)^x. Then y = u + v → dy/dx = du/dx + dv/dx.
For u = x^{sin x}:
ln u = sin x ⋅ ln x → (1/u)du/dx = cos x ln x + sin x/x.
du/dx = x^{sin x}[cos x ln x + sin x/x].
For v = (sin x)^x:
ln v = x ln(sin x) → (1/v)dv/dx = ln(sin x) + x cot x.
dv/dx = (sin x)^x[ln(sin x) + x cot x].
dy/dx = x^{sin x}[cos x ln x + sin x/x] + (sin x)^x[ln(sin x) + x cot x].

🤯 T5. If y = x^{x^x} (x to the power x to the power x), find dy/dx.

Let z = x^x so y = x^z.
ln y = z ln x = x^x ln x.
(1/y) dy/dx = d/dx[x^x ln x].
= ln x ⋅ d/dx(x^x) + x^x ⋅ d/dx(ln x)
= ln x ⋅ x^x(1+ln x) + x^x ⋅ (1/x)
= x^x[ln x(1+ln x) + 1/x].
dy/dx = y ⋅ x^x[ln x + ln²x + 1/x] = x^{x^x} ⋅ x^x[ln x(1+ln x) + 1/x].

6. Higher-Order Derivatives

6.1

Second Derivative & n^th Order Derivatives

Differentiate the derivative again to get the second derivative

⚡ Higher Derivatives — Notation & Method

Second derivative: d²y/dx² = d/dx(dy/dx) = f′′(x) = y′′ (differentiate dy/dx once more) Third derivative: d³y/dx³ = f′′′(x) (differentiate again) General n-th derivative: dⁿy/dxⁿ = f^(n)(x) Standard n-th derivatives (important for NDA): dⁿ/dxⁿ (xⁿ) = n! (constant after n differentiations) dⁿ/dxⁿ (eˣ) = eˣ (unchanged under any order) dⁿ/dxⁿ (sin x) = sin(x + nπ/2) dⁿ/dxⁿ (cos x) = cos(x + nπ/2) dⁿ/dxⁿ (aˣ) = aˣ (ln a)ⁿ Physical meanings of second derivative: In kinematics: d²s/dt² = acceleration In geometry: f′′(x) > 0 → curve is concave up (cup shape) f′′(x) < 0 → curve is concave down (cap shape)

The second derivative test for maxima/minima: if f′(a)=0 and f′′(a)>0 → local minimum. If f′′(a)<0 → local maximum. This is tested in application questions in NDA.

Worked Example — Second Derivative

Find d²y/dx² if y = x&sup4; − 3x² + 5.

dy/dx = 4x³ − 6x.

d²y/dx² = d/dx(4x³ − 6x) = 12x² − 6 = 6(2x² − 1).

Worked Example — Verifying a Differential Equation

If y = A sin x + B cos x, show that d²y/dx² + y = 0.

dy/dx = A cos x − B sin x.

d²y/dx² = −A sin x − B cos x = −(A sin x + B cos x) = −y.

Therefore d²y/dx² + y = −y + y = 0 ✓.

📝 TOPIC-WISE PYQ

Higher-Order Derivatives — NDA-Pattern Questions

Q13. If y = e^2x, find d²y/dx².

(a) 2e^2x (b) 4e^2x (c) e^2x (d) 8e^2x

Answer: (b) 4e^2x
dy/dx = 2e^2x. d²y/dx² = 2 ⋅ 2e^2x = 4e^2x.

Q14. If y = x e^x, find d²y/dx².

(a) e^x(x+1) (b) e^x(x+2) (c) xe^x (d) 2e^x

Answer: (b) e^x(x+2)
dy/dx = e^x + xe^x = e^x(1+x).
d²y/dx² = e^x(1+x) + e^x(1) = e^x(2+x) = e^x(x+2).

Q15. If y = sin x, find the 10^th derivative d¹⁰y/dx¹⁰.

(a) −sin x (b) sin x (c) cos x (d) −cos x

Answer: (b) sin x
dⁿ/dxⁿ(sin x) = sin(x + nπ/2).
For n=10: sin(x + 10π/2) = sin(x + 5π) = sin(x + 4π + π) = sin(x+π) …
Actually: 10π/2 = 5π. sin(x+5π) = sin(x+4π+π) = sin(x+π) = −sin x.
Recheck: n=1: cos x; n=2: −sin x; n=3: −cos x; n=4: sin x (cycle of 4). 10 = 4×2+2, so same as n=2 → −sin x. Answer: (a).

🔥 TRICKY QUESTIONS

Higher-Order Derivatives — Verification Traps

🤯 T6. If y = e^ax sin(bx), find d²y/dx² and verify the equation d²y/dx² − 2a(dy/dx) + (a²+b²)y = 0.

dy/dx = ae^ax sin(bx) + be^ax cos(bx) = e^ax[a sin(bx) + b cos(bx)].
d²y/dx² = a⋅e^ax[a sin(bx)+b cos(bx)] + e^ax[ab cos(bx)−b² sin(bx)]
= e^ax[a² sin(bx) + 2ab cos(bx) − b² sin(bx)]
= e^ax[(a²−b²)sin(bx) + 2ab cos(bx)].
Now: d²y/dx² − 2a(dy/dx) + (a²+b²)y
= e^ax[(a²−b²)sin+2ab cos] − 2ae^ax[a sin+b cos] + (a²+b²)e^ax sin
= e^ax[(a²−b²−2a²+a²+b²)sin + (2ab−2ab)cos] = e^ax[0⋅sin + 0] = 0 ✓.

📝 Master Formula Sheet — MN12 Differentiation

All critical formulae for rapid pre-exam revision.

∂ Standard Derivatives

d/dx(xⁿ) = nx^(n-1)
d/dx(sin x) = cos x
d/dx(cos x) = −sin x
d/dx(tan x) = sec² x
d/dx(eˣ) = eˣ
d/dx(aˣ) = aˣ ln a
d/dx(ln x) = 1/x

➕ Sum & ✖ Product Rules

d/dx[u+v] = u′+v′
d/dx[uv] = u′v + uv′
Memory: "first ⋅ d(second) + second ⋅ d(first)"
d/dx[cu] = c⋅u′ (constant factor)

÷ Quotient & 🔗 Chain Rules

d/dx[u/v] = (u′v − uv′) / v²
d/dx[f(g)] = f′(g) ⋅ g′
Chain: outer′ × inner′
d/dx[e^g(x)] = e^g(x) ⋅ g′(x)
d/dx[ln(g(x))] = g′(x)/g(x)

∞ Implicit & Parametric

d/dx(yⁿ) = ny^(n−1) dy/dx
d/dx(xy) = x dy/dx + y
Collect dy/dx terms, factor, solve
Parametric: dy/dx = (dy/dt)/(dx/dt)

🔗 Log Differentiation

For y=[f(x)]^[g(x)]: take ln both sides
d/dx(xˣ) = xˣ(1+ln x)
(1/y) dy/dx → then multiply by y
Products: ln(uvw) = ln u+ln v+ln w

∛ Higher-Order Derivatives

d²y/dx² = d/dx(dy/dx)
dⁿ(eˣ)/dxⁿ = eˣ
dⁿ(sin x)/dxⁿ = sin(x+nπ/2)
dⁿ(xⁿ)/dxⁿ = n!
f′′>0: concave up; f′′<0: concave down

⚡ Quick Revision Booster — MN12 Differentiation

∂ Standard Quick-Ref

d(xⁿ) = nx^(n−1) ← always
d(sin) = cos, d(cos) = −sin
d(eˣ) = eˣ (unique!)
d(ln x) = 1/x
d(aˣ) = aˣ ln a

✖ Product Rule

(uv)′ = u′v + uv′
Tip: write u, v; find u′, v′; add crossed
(uvw)′ = u′vw + uv′w + uvw′
x²eˣ: (2x)eˣ + x²(eˣ) = eˣ(x²+2x)

🔗 Chain Rule

Identify: outer f, inner g
d/dx = f′(g) × g′
sin(3x): cos(3x) × 3 = 3cos(3x)
e^x²: e^x² × 2x = 2xe^x²
ln(cos x): (1/cos x)×(−sin x) = −tan x

∞ Implicit & Parametric

Every y-term gets × dy/dx
d(y²)/dx = 2y dy/dx
Collect, factor, solve for dy/dx
Parametric: (dy/dt) ÷ (dx/dt)
x=t², y=t³: dy/dx = 3t²/2t = 3t/2

🔗 Log Diff. Steps

y=xˣ → take ln: ln y = x ln x
Differentiate: (1/y)y′ = …
Multiply by y: y′ = y(…)
d(xˣ)/dx = xˣ(1+ln x)
Use when exponent has x

🚨 Critical Exam Traps

d(xˣ)/dx ≠ x⋅x^(x−1) (wrong!)
d(cos x)/dx = −sin x (NOT +sin x)
Quotient: denominator is v², sign is minus
x² sin x → product rule (NOT 2x cos x)
Implicit: d(y)/dx = dy/dx, not 1
Cycle for sin: n mod 4 determines result

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