Approximation using Differentials
Applications of Derivatives
15
JEE Qs
8%
Hard
45
min
Master the skill of judiciously choosing the base value 'x' and the increment 'Δx' to simplify calculations while maintaining approximation accuracy.
🧮 Key Formulas
✅ Key Points for JEE
- 1The core idea is that for a very small change in x (Δx), the actual change in y (Δy) is approximately equal to the differential of y (dy).
- 2The approximation formula f(x + Δx) ≈ f(x) + f'(x)Δx is derived from dy/dx ≈ Δy/Δx.
- 3Always identify 'x' as a value near the desired input for which f(x) is easily calculable, and 'Δx' as the small difference between the desired input and 'x'.
- 4Ensure Δx is sufficiently small for the approximation to be reasonably accurate; larger Δx values lead to greater errors.
- 5The differential dy represents the change in y along the tangent line at x, while Δy is the actual change in y along the curve.
⚠️ Common Mistakes
- ✕Confusing the actual change in y (Δy) with the approximate change in y (dy).
- ✕Incorrectly identifying or choosing the base value 'x' and the increment 'Δx'. Often students pick 'x' and 'Δx' such that 'x' is not easily computable or 'Δx' is too large.
- ✕Errors in calculating the derivative f'(x) or substituting values into the approximation formula.
📝 Practice Questions
See allQ10.Let the function f(x) = (x2 + 1) x2 −ax + 2 + cos |x| be not differentiable at the two points x = α = 2 and x = β . Then the distance of the point (α, β) from the line 12x + 5y + 10 = 0 is equal to : (1) 5 (2) 4 (3) 3 (4) 2
Q24.Let the function, f(x) = {−3ax2a2 + bx,−2, xx <⩾11 be differentiable for all x ∈R, where a > 1, b ∈R. If the area of the region enclosed by y = f(x) and the line y = −20 is α + β√3, α, β ∈Z , then the value of α + β is ________
Q8. Let f(x) = ∫x20 t2−8t+15et dt, respectively, are : (1) 2 and 3 (2) 2 and 2 (3) 3 and 2 (4) 1 and 3
Q23.If the set of all values of a, for which the equation 5x3 −15x −a = 0 has three distinct real roots, is the interval (α, β), then β −2α is equal to ______
Q13.A spherical chocolate ball has a layer of ice-cream of uniform thickness around it. When the thickness of the ice-cream layer is 1 cm , the ice-cream melts at the rate of 81 cm3/min and the thickness of the ice-cream layer decreases at the rate of 1 cm/min. The surface area (in cm2 ) of the chocolate ball (without the ice- 4π cream layer) is : (1) 196π (2) 256π (3) 225π (4) 128π
Q20.Let →a = ^i + 2^j + 3^k,→b = 3^i + ^j −^k and →c be three vectors such that →c is coplanar with →a and →b. If the vector →C is perpendicular to →b and →a ⋅→c = 5, then |→c| is equal to (1) √116 (2) 3√21 (3) 16 (4) 18
NCERT Chapters
- Class 12 Maths Ch 6: Application of Derivatives (Section: Differentials)