Implicit Differentiation
Differentiation
2
JEE Qs
8%
Hard
60
min
Master the application of the chain rule to 'y' terms and practice meticulous algebraic manipulation to correctly isolate dy/dx.
🧮 Key Formulas
✅ Key Points for JEE
- 1When differentiating an equation implicitly, treat 'y' as an unknown function of 'x' (i.e., y = f(x)).
- 2Differentiate both sides of the equation with respect to 'x'.
- 3Apply the chain rule whenever differentiating a term involving 'y'; always multiply its derivative by dy/dx.
- 4After differentiating, rearrange the equation algebraically to isolate dy/dx.
- 5Remember to apply product rule (d/dx(uv) = u'v + uv') for terms involving products of x and y (e.g., xy) and quotient rule if necessary.
⚠️ Common Mistakes
- ✕Forgetting to multiply by dy/dx when differentiating a term containing 'y' with respect to 'x'.
- ✕Incorrectly applying the product or quotient rule for terms involving both 'x' and 'y' (e.g., d/dx(x*y)).
- ✕Algebraic errors in collecting terms and isolating dy/dx.
- ✕Treating 'y' as a constant or 'x' as a constant during differentiation, instead of an implicit function.
📝 Practice Questions
See allQ71.Let f(x) = x5 + 2x3 + 3x + 1, x ∈R , and g(x) be a function such that g(f(x)) = x for all x ∈R . Then g(7) g′(7) is equal to : (1) 14 (2) 42 (3) 7 (4) 1
Q73.If f(x) = {x30 sin, x (= 0 (1) f ′′ ( π2 ) = 24−π22π (2) f ′′ ( π2 ) = 12−π22π (3) f ′′(0) = 1 (4) f ′′(0) = 0
Q73.Let 𝑓𝑥= 2𝑥2 + 5𝑥- 3, 𝑥∈𝑅. If 𝑚 and 𝑛 denote the number of points where 𝑓 is not continuous and not differentiable respectively, then 𝑚+ 𝑛 is equal to: (1) 5 (2) 2 (3) 0 (4) 3
Q74.Let f(x) = x5 + 2ex/4 for all x ∈R. Consider a function g(x) such that (g ∘f)(x) = x for all x ∈R. Then the value of 8g′(2) is : (1) 2 (2) 8 (3) 4 (4) 16 is equal to :
Q74.Let β(m, n) = ∫10 xm−1(1 −x)n−1 dx, m, n > 0 . If ∫10 (1 −x10) dx = a × β(b, c), then 100(a + b + c) equals____ (1) 1021 (2) 2120 (3) 2012 (4) 1120 JEE Main 2024 (05 Apr Shift 2) JEE Main Previous Year Paper
Q72.Suppose for a differentiable function h, h(0) = 0, h(1) = 1 and h′(0) = h′(1) = 2. If g(x) = h (ex)eh(x) , then g′(0) is equal to: (1) 5 (2) 4 (3) 8 (4) 3
NCERT Chapters
- Class 12 Mathematics Ch 5: Continuity and Differentiability