Chapter 5: Continuity and Differentiability
Key Rules:
• Chain Rule: d/dx[f(g(x))] = f'(g(x)) · g'(x)
• Product Rule: d/dx(uv) = uv' + vu'
• Quotient Rule: d/dx(u/v) = (vu' − uv')/v²
• If y = eax, dy/dx = aeax
Q1. Find dy/dx if y = e3x.
Solution
dy/dx = 3e3x
Using chain rule: d/dx(e3x) = e3x · d/dx(3x) = 3e3x. ✓
Q2. Find dy/dx if y = log(sin x).
Solution
dy/dx = (1/sin x) · cos x = cos x/sin x = cot x
Q3. Find dy/dx if y = (sin x)cos x.
Solution
Take log on both sides: log y = cos x · log(sin x)
Differentiate both sides:
(1/y)(dy/dx) = −sin x · log(sin x) + cos x · (cos x/sin x)
dy/dx = y [−sin x · log(sin x) + cos²x/sin x]
= (sin x)cos x [cos x · cot x − sin x · log(sin x)]
Q4. Verify that y = ex sin x satisfies the equation d²y/dx² − 2 dy/dx + 2y = 0.
Solution
y = ex sin x
dy/dx = ex sin x + ex cos x = ex(sin x + cos x)
d²y/dx² = ex(sin x + cos x) + ex(cos x − sin x) = 2ex cos x
LHS = 2ex cos x − 2ex(sin x + cos x) + 2ex sin x
= 2ex cos x − 2ex sin x − 2ex cos x + 2ex sin x = 0 ✓
Q5. Find dy/dx if x² + y² = 25 (implicit differentiation).
Solution
Differentiating both sides w.r.t. x:
2x + 2y(dy/dx) = 0
2y(dy/dx) = −2x
dy/dx = −x/y
Q6. Check the continuity of f(x) = |x| at x = 0.
Solution
f(x) = { −x, if x < 0
{ x, if x ≥ 0
f(0) = 0
LHL = limx→0− (−x) = 0
RHL = limx→0+ (x) = 0
LHL = RHL = f(0) = 0
∴ f is continuous at x = 0. ✓