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Ex 5.6, 3 - Chapter 5 Class 12 Continuity and Differentiability
Last updated at Dec. 16, 2024 by Teachoo
Derivatives in parametric form
Chapter 5 Class 12 Continuity and Differentiability
Concept wise
- Checking continuity at a given point
- Checking continuity at any point
- Checking continuity using LHL and RHL
- Algebra of continous functions
- Continuity of composite functions
- Checking if funciton is differentiable
- Finding derivative of a function by chain rule
- Finding derivative of Implicit functions
- Finding derivative of Inverse trigonometric functions
- Finding derivative of Exponential & logarithm functions
- Logarithmic Differentiation - Type 1
- Logarithmic Differentiation - Type 2
-
Derivatives in parametric form
- Finding second order derivatives - Normal form
- Finding second order derivatives- Implicit form
- Proofs
- Verify Rolles theorem
- Verify Mean Value Theorem
Transcript
Ex 5.6, 3 If x and y are connected parametrically by the equations without eliminating the parameter, Find ππ¦/ππ₯, π₯=sinβ‘π‘, π¦=cosβ‘2π‘Here, ππ¦/ππ₯ = (ππ¦/ππ‘)/(ππ₯/ππ‘) Calculating π π/π π ππ¦/ππ‘ " " = π(cosβ‘2π‘)/ππ‘ ππ¦/ππ‘ = βsinβ‘2π‘ . 2 ππ¦/ππ‘ = β2 sinβ‘2π‘ Calculating π π/π π ππ₯/ππ‘ " " = π(sinβ‘π‘ )/ππ‘ ππ₯/ππ‘ " " = cosβ‘π‘ Therefore ππ¦/ππ₯ = (ππ¦/ππ‘)/(ππ₯/ππ‘) ππ¦/ππ₯ = (β2 sinβ‘2π‘" " )/cosβ‘π‘ ππ¦/ππ₯ = (β2(2 sinβ‘π‘ .γ cosγβ‘π‘ ))/cosβ‘π‘ π π/π π = βπ πππβ‘π (As sin 2ΞΈ = 2 sin ΞΈ cos ΞΈ)
