Concept of Continuity
A function is said to be continuous at a point if you can draw its graph through that point without lifting your pen. In AAI ATC exam, this topic tests your understanding of limits and function behavior.
A function f is continuous at x = c if:
i.e., Left Hand Limit = Right Hand Limit = Value of function at c
[ ADD IMAGE HERE ] – Graph showing continuous vs. discontinuous functions (e.g., jump discontinuity, removable discontinuity)
Three conditions for continuity at x = c:
1. f(c) must be defined 2. lim(x→c) f(x) must exist 3. Both must be equal
Every polynomial and rational function is continuous in its domain. Trigonometric functions like sin x and cos x are continuous everywhere. tan x is discontinuous at x = (2n+1)π/2.
lim(x→2) (x²–4)/(x–2) = lim(x→2)(x+2)(x–2)/(x–2) = lim(x→2)(x+2) = 4. For continuity, k = 4.
For x < 3, |x–3| = 3–x. As x→3⁻, f(x) = 3–x → 0. Hence LHL = 0 = f(3).
For continuity at x=2: lim(x→2⁻) kx² = lim(x→2⁺) 3 → k(4) = 3 → k = 3/4.
Algebra of Continuous Functions
If f and g are two continuous functions at x = c, then their algebraic combinations are also continuous. This theorem is frequently tested in AAI ATC exams.
Theorem 1: If f and g are continuous at x = c, then:
- f + g is continuous at x = c
- f – g is continuous at x = c
- f · g is continuous at x = c
- f/g is continuous at x = c (provided g(c) ≠ 0)
[ ADD IMAGE HERE ] – Venn diagram or table showing continuous function operations
✅ Polynomial functions — continuous everywhere
✅ Rational functions — continuous except where denominator = 0
✅ sin x, cos x — continuous everywhere
✅ tan x — discontinuous at x = (2n+1)π/2
✅ |x| — continuous everywhere
f(x) is discontinuous where x²–4 = 0, i.e., x = 2 and x = –2. Hence 2 points of discontinuity.
Using standard limit: lim(x→0) sin(x)/x = 1 = f(0). All three conditions of continuity satisfied.
g(x) = 1/x is undefined at x = 0. cos is continuous everywhere, so the composite cos(1/x) is continuous for x ≠ 0.
Differentiability of Functions
A function is differentiable at x = c if its derivative exists at that point — i.e., LHD = RHD. This is a stricter condition than continuity.
The derivative of f at c is:
f is differentiable at c ⟺ LHD = RHD (both finite)
Theorem 3: If f is differentiable at c, then f is continuous at c.
⚠️ Converse is NOT always true: f(x) = |x| is continuous at 0 but NOT differentiable at 0.
[ ADD IMAGE HERE ] – Graph of f(x) = |x| showing sharp corner at x = 0 (not differentiable)
For f(x) = |x|: LHD at 0 = –1 and RHD at 0 = +1. Since LHD ≠ RHD, f is not differentiable at x = 0, even though it is continuous there.
f'(0) = lim(h→0) f(h)/h = lim(h→0) h|h|/h = lim(h→0)|h| = 0. Both LHD and RHD = 0.
f(0) = 0. lim(x→0) x²sin(1/x) = 0 (bounded × 0). f'(0) = lim(h→0) h²sin(1/h)/h = lim(h→0) h·sin(1/h) = 0.
[x] is discontinuous (and hence not differentiable) at every integer value. It has a jump discontinuity at each integer.
Chain Rule & Composite Functions
The Chain Rule is used to differentiate composite functions. It's one of the most frequently tested tools in AAI ATC mathematics.
If f = v ∘ u, and t = u(x), then:
[ ADD IMAGE HERE ] – Flow diagram showing Chain Rule: x → u(x) → v(u(x)) with derivative arrows
Example: Find d/dx [sin(x²)]
For d/dx[f(ax+b)ⁿ] = n·f(ax+b)ⁿ⁻¹ · f'(ax+b) · a. Always multiply by derivative of the inner function!
Let t = x³. d/dt[cos t] = –sin t. dt/dx = 3x². So d/dx[cos(x³)] = –sin(x³) · 3x² = –3x² sin(x³).
dy/dx = 5(2x+1)⁴ · 2 = 10(2x+1)⁴. At x=0: 10(1)⁴ = 10.
d/dx[sin²(3x)] = 2sin(3x)·cos(3x)·3 = 3·2sin(3x)cos(3x) = 3sin(6x).
Implicit Differentiation
When y cannot be expressed explicitly as a function of x, we differentiate both sides with respect to x, treating y as a function of x and using chain rule.
For implicit functions F(x, y) = 0, differentiate both sides w.r.t. x:
Key: d/dx[f(y)] = f'(y) · dy/dx
[ ADD IMAGE HERE ] – Example: Circle x² + y² = r² with tangent line showing implicit diff.
Example: Find dy/dx if x² + y² = 25
Differentiate: 3x² + 3y²(dy/dx) = 3y + 3x(dy/dx). At (1,1): 3 + 3(dy/dx) = 3 + 3(dy/dx) → rearranging: (3y²–3x)(dy/dx) = 3y–3x² → dy/dx = (y–x²)/(y²–x) = 0/0 — use L'Hôpital or factor: = –1.
cos(x+y)·(1+dy/dx) = 2y·(dy/dx). cos(x+y) = dy/dx·[2y–cos(x+y)]. dy/dx = cos(x+y)/[2y–cos(x+y)].
x² + 2xy + y² = (x+y)² = 9. Differentiating: 2(x+y)(1+dy/dx) = 0. So 1+dy/dx = 0 → dy/dx = –1.
Inverse Trig Derivatives
Derivatives of inverse trigonometric functions are directly asked in AAI ATC exams. These are derived using the chain rule on inverse function identities.
| Function f(x) | Domain | Derivative f'(x) |
|---|---|---|
| sin⁻¹(x) | (–1, 1) | 1/√(1–x²) |
| cos⁻¹(x) | (–1, 1) | –1/√(1–x²) |
| tan⁻¹(x) | ℝ | 1/(1+x²) |
| cot⁻¹(x) | ℝ | –1/(1+x²) |
| sec⁻¹(x) | |x|>1 | 1/(|x|√(x²–1)) |
| cosec⁻¹(x) | |x|>1 | –1/(|x|√(x²–1)) |
[ ADD IMAGE HERE ] – Graphs of sin⁻¹x and cos⁻¹x showing domain and range
Put x = sinθ. Then 2x√(1–x²) = 2sinθcosθ = sin2θ. So sin⁻¹(sin2θ) = 2θ = 2sin⁻¹x. d/dx[2sin⁻¹x] = 2/√(1–x²).
tan⁻¹((1–x)/(1+x)) = tan⁻¹(1) – tan⁻¹(x) = π/4 – tan⁻¹x. Derivative = –1/(1+x²).
Put x = sinθ: 1–2sin²θ = cos2θ. cos⁻¹(cos2θ) = 2θ = 2sin⁻¹x. Derivative = 2/√(1–x²).
Exponential & Logarithmic Functions
Exponential and logarithmic functions have elegant differentiation rules. The natural exponential function eˣ is unique — it is its own derivative!
Theorem 5: Standard Derivatives
| Function | Derivative |
|---|---|
| eˣ | eˣ |
| e^(f(x)) | e^(f(x)) · f'(x) |
| log x | 1/x |
| log(f(x)) | f'(x)/f(x) |
| aˣ (a > 0) | aˣ log a |
[ ADD IMAGE HERE ] – Graph of y = eˣ and y = ln(x) showing mirror symmetry about y=x
d/dx[e^(sin x)] = e^(sin x) · d/dx[sin x] = e^(sin x) · cos x.
Using chain rule: d/dx[log(f(x))] = f'(x)/f(x) = 2x/(x²+3).
d/dx[aˣ] = aˣ · log a. So d/dx[2ˣ] = 2ˣ · log 2.
Logarithmic Differentiation
Used to differentiate functions of the form y = [u(x)]^v(x) where both base and exponent are functions of x. Take log of both sides first!
For y = [u(x)]^v(x), take log: log y = v(x) · log[u(x)]
[ ADD IMAGE HERE ] – Step-by-step worked example of d/dx [xˢⁱⁿˣ]
Example: Differentiate xˢⁱⁿˣ
log y = x log x. (1/y)(dy/dx) = log x + 1 = 1 + log x. dy/dx = xˣ(1 + log x).
Let y = (log x)ˣ. log y = x · log(log x). (1/y)(dy/dx) = log(log x) + x·(1/(log x))·(1/x) = log(log x) + 1/log x. Multiply by y.
At x=π/2: sin(π/2)=1, cos(π/2)=0. y = 1⁰ = 1. dy/dx = y[–sin x · log(sin x) + cos x · cos x/sin x]. At x=π/2: y=1, sin x=1, cos x=0, log 1=0. So dy/dx = 1·[0 + 0] = 0.
Parametric Differentiation
When x and y are both expressed in terms of a third variable (parameter) t, we use:
[ ADD IMAGE HERE ] – Parametric curve (circle: x=cosθ, y=sinθ) with tangent line
• Parabola: x = at², y = 2at → dy/dx = 1/t
• Circle: x = a cosθ, y = a sinθ → dy/dx = –cot θ
• Cycloid: x = a(θ–sinθ), y = a(1–cosθ) → dy/dx = tan(θ/2)
dx/dt = 2t, dy/dt = 2. dy/dx = 2/2t = 1/t.
dx/dθ = –a sinθ, dy/dθ = b cosθ. dy/dx = b cosθ / (–a sinθ) = –(b/a) cotθ.
dx/dθ = –3sinθ+3cos²θsinθ = –3sinθcos²θ (after factor). dy/dθ = 3cosθ–3sin²θcosθ = 3cos³θ. dy/dx = –cos³θ/sin³θ = –cot³θ. At θ=π/4: –(1)³ = –1.
Second Order Derivatives
The second derivative is the derivative of the first derivative. It is denoted as d²y/dx², f''(x), or y₂. Used extensively in AAI ATC applied problems.
Physical meaning: rate of change of the rate of change (acceleration in physics)
[ ADD IMAGE HERE ] – Graph showing f, f', f'' with concavity interpretation
• y = sin x → y'' = –sin x = –y
• y = cos x → y'' = –cos x = –y
• y = eˣ → y'' = eˣ = y
• y = xⁿ → y'' = n(n–1)x^(n–2)
dy/dx = 3x² + 3. d²y/dx² = 6x. At x = 1: 6(1) = 6.
This is a standard result (Theorem in textbook). y'' = Am²e^(mx)+Bn²e^(nx). y' = Ame^(mx)+Bne^(nx). Substituting: y''–(m+n)y'+mny = 0.
y = log x. y' = 1/x. y'' = –1/x². x²y'' + xy' = x²(–1/x²) + x(1/x) = –1 + 1 = 0.
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