Lesson 13 of 13 · 14 min

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Must-know facts

16 facts

1Continuous at c: LHL = RHL = f(c), all finite.
2A point of discontinuity must be in the domain; 1/x is a continuous function.
3[x] is discontinuous at every integer: LHL = c − 1, RHL = c.
4Sums, differences, products, quotients (denominator ≠ 0) and composites of continuous functions are continuous.
5Polynomials, sin, cos, |x|, eˣ and log x (x > 0) are continuous on their domains.
6Differentiable at c ⇒ continuous at c; the converse is false (|x| at 0).
7Differentiable at c: left and right derivatives finite and equal.
8Chain rule: d/dx v(u(x)) = v′(u(x))·u′(x).
9Implicit: every y term gets a factor dy/dx.
10(sin⁻¹x)′ = 1/√(1 − x²), (cos⁻¹x)′ = −1/√(1 − x²), (tan⁻¹x)′ = 1/(1 + x²).
11(eˣ)′ = eˣ, (log x)′ = 1/x, (aˣ)′ = aˣ log a.
12For u^v, take logs first; (xˣ)′ = xˣ(1 + log x).
13Parametric: dy/dx = (dy/dt)/(dx/dt), dx/dt ≠ 0.
14y = A sin x + B cos x satisfies y″ + y = 0.
15Rolle: f(a) = f(b) ⇒ some f′(c) = 0. MVT: some f′(c) equals the chord slope (JEE).
16kx + 1 / 3x − 5 joined at 5 is continuous for k = 9/5.

Common traps

Where marks are lost

Calling 1/x discontinuous at 0.

0 is not in the domain, so it is not tested. 1/x is continuous at every point where it is defined.

Checking only that the two one-sided limits agree.

The value f(c) must match them too. x³ + 3 with f(0) = 1 has LHL = RHL = 3 but is discontinuous at 0.

Assuming continuous means differentiable.

|x| is continuous at 0 with left derivative −1 and right derivative +1, so it has no derivative there.

Dropping the inner derivative: d/dx sin(x²) = cos(x²).

Multiply by the derivative of x²: the answer is 2x cos(x²).

Treating y as a constant in implicit differentiation.

d/dx(y²) = 2y dy/dx and d/dx(sin y) = cos y dy/dx.

Using the power rule on xˣ to get x·xˣ⁻¹.

The exponent varies, so take logs: d/dx(xˣ) = xˣ(1 + log x).

Writing d/dx(aˣ) = x aˣ⁻¹.

It is aˣ log a; the power rule needs a constant exponent.

Finding the parametric second derivative as (d²y/dt²)/(d²x/dt²).

Differentiate dy/dx with respect to t and divide by dx/dt.

Applying Rolle's theorem without checking differentiability.

|x| on [−1, 1] has f(−1) = f(1) but no horizontal tangent; the corner at 0 breaks the hypothesis.

Formulas

9 to know

Continuity at c

lim(x→c⁻) f = lim(x→c⁺) f = f(c)

All three must exist and agree.

Derivative at c

f′(c) = lim(h→0) [f(c + h) − f(c)]/h

Left and right versions must be finite and equal.

Chain rule

df/dx = (dv/dt)·(dt/dx), with t = u(x)

Extends to any number of links.

Inverse trig

(sin⁻¹x)′ = 1/√(1 − x²), (cos⁻¹x)′ = −1/√(1 − x²), (tan⁻¹x)′ = 1/(1 + x²)

First two on (−1, 1); the third on R.

Exponential and log

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

log x needs x > 0.

Logarithmic differentiation

y = u^v ⇒ y′/y = v′ log u + v u′/u

Needs u > 0.

Parametric

dy/dx = (dy/dt)/(dx/dt)

Valid where dx/dt ≠ 0.

Second derivative

d²y/dx² = d/dx(dy/dx)

Also written y″ or y₂.

Mean value theorem (JEE)

f′(c) = [f(b) − f(a)]/(b − a), a < c < b

f continuous on [a, b], differentiable on (a, b); Rolle when f(a) = f(b).

Key terms

6 terms

Continuous at c: The limit of f at c exists and equals f(c).
Point of discontinuity: A point of the domain at which f is not continuous.
Differentiable at c: The left-hand and right-hand derivatives at c are finite and equal.
Implicit function: y given by a relation with x that is not solved for y.
Parameter: A third variable through which both x and y are expressed.
Second order derivative: The derivative of the derivative, d²y/dx².