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Ex 5.3, 5 - Chapter 5 Class 12 Continuity and Differentiability (Term 1)
Last updated at March 10, 2021 by Teachoo
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 5.3, 5 Find ππ¦/ππ₯ in, π₯2 + π₯π¦ + π¦2 = 100 π₯2 + π₯π¦ + π¦2 = 100 Differentiating both sides π€.π.π‘.π₯ . π(π₯2 + π₯π¦ + π¦2)/ππ₯ = (π (100))/ππ₯ π(π₯2)/ππ₯ + π(π₯π¦)/ππ₯ + (π(π¦2))/ππ₯ = (π (100))/ππ₯ 2π₯^(2β1) + (π(π₯π¦))/ππ₯ + (π(π¦2))/ππ₯ Γ ππ¦/ππ¦ = 0 2π₯ + (π(π₯π¦))/ππ₯ + (π(π¦2))/ππ¦ Γ ππ¦/ππ₯ = 0 As (π₯^π )^β²=ππ₯^(πβ1) & derivative of a constant is zero 2π₯ + (π(π₯π¦))/ππ₯ + 2π¦^(2β1) . ππ¦/ππ₯ = 0 2π₯ + (π(π₯π¦))/ππ₯ + 2π¦ . ππ¦/ππ₯ = 0 Using product rule in π₯π¦ = π₯^β² π¦+π¦^β² π₯ 2π₯ + ((π(π₯))/ππ₯ .π¦+π(π¦)/ππ₯.π₯) + 2π¦ . ππ¦/ππ₯ = 0 2π₯ + (1 .π¦+π₯.π(π¦)/ππ₯) + 2π¦ . ππ¦/ππ₯ = 0 2π₯ + π¦ + π₯ . ππ¦/ππ₯ + 2π¦ . ππ¦/ππ₯ = 0 (2π₯ + π¦) + ππ¦/ππ₯ . (π₯ + 2π¦) = 0 ππ¦/ππ₯ . (π₯ + 2π¦) = β (2π₯ + π¦) π π/π π = (β( ππ + π ))/( ( π + ππ ))
