Q1. Differentiate y = sin(x²) using the chain rule.
Let u = x², so y = sin(u)
dy/du = cos(u), du/dx = 2x
dy/dx = cos(x²) × 2x = 2x cos(x²)
Continuity and Differentiability — Class 12
A function is continuous if you can draw it without lifting your pen. Learn the exact test, then the chain rule that unlocks composite derivatives.
Continuity at a pointf is continuous at x = a if lim(x→a) f(x) = f(a) — the left-hand limit, right-hand limit and function value all agree.
Differentiabilityf is differentiable at a if its derivative exists there. Differentiable ⇒ continuous, but NOT the reverse (e.g. |x| at x=0).
Chain ruleFor y = f(g(x)): dy/dx = f'(g(x)) · g'(x) — differentiate the outer function, then multiply by the derivative of the inner.
Solved examples
Q2. Is f(x) = |x| differentiable at x = 0?
Left-hand derivative at 0: slope of −x is −1
Right-hand derivative at 0: slope of x is +1
Since −1 ≠ 1, f is NOT differentiable at x = 0 (though it IS continuous there).
Formula sheet
| Quantity | Formula |
|---|---|
| Continuity condition | LHL = RHL = f(a) |
| Chain rule | dy/dx = dy/du × du/dx |
| Derivative of eˣ | d/dx(eˣ) = eˣ |
| Derivative of ln x | d/dx(ln x) = 1/x |
Common mistakes & exam wins
- Every differentiable function is continuous, but a continuous function need not be differentiable (classic example: |x| at x = 0).
- The chain rule is needed whenever you differentiate a "function of a function" — spot the inner function first.
- For implicit functions (like x² + y² = 25), differentiate both sides with respect to x and solve for dy/dx.
Frequently asked questions
What is the condition for a function to be continuous at a point?
The left-hand limit, right-hand limit, and the function value at that point must all be equal.
What is the chain rule?
For a composite function y = f(g(x)), the derivative is dy/dx = f'(g(x)) · g'(x).
Is every continuous function differentiable?
No — a function can be continuous but not differentiable at a point, like |x| at x = 0.