Example 35 - Chapter 5 Class 12 Continuity and Differentiability (Term 1)
Last updated at April 13, 2021 by Teachoo
Derivatives in parametric form
Chapter 5 Class 12 Continuity and Differentiability (Term 1)
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
Example 35 Find ππ¦/ππ₯ , if π₯ = ππ‘2, π¦ = 2ππ‘ . ππ¦/ππ₯ = ππ¦/ππ₯ Γ ππ‘/ππ‘ ππ¦/ππ₯ = ππ¦/ππ‘ Γ ππ‘/ππ₯ ππ¦/ππ₯ = (ππ¦/ππ‘)/(ππ₯/ππ‘) Calculating π π/π π π¦ = 2ππ‘" " ππ¦/ππ‘ " " = π(2ππ‘)/ππ‘ ππ¦/ππ‘ = 2π . π(π‘)/ππ‘ ππ¦/ππ‘ = ππ Calculating π π/π π π₯=ππ‘2 ππ₯/ππ‘ " " = π(ππ‘2)/ππ‘ ππ₯/ππ‘ " " = π . π(π‘2)/ππ‘ " " ππ₯/ππ‘ " " = πππ Therefore ππ¦/ππ₯ = (ππ¦/ππ‘)/(ππ₯/ππ‘) ππ¦/ππ₯ = 2π/2aπ‘ π π/π π = π/π
