Ex 5.3, 15 - Chapter 5 Class 12 Continuity and Differentiability
Last updated at Dec. 16, 2024 by Teachoo
Finding derivative of Inverse trigonometric functions
Derivative of cos-1 x (Cos inverse x)
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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
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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, 15 Find 𝑑𝑦/𝑑𝑥 in, y = sec–1 (1/( 2𝑥2−1 )), 0 < x < 1/√2 y = sec–1 (1/( 2𝑥^2 − 1 )) 𝒔𝒆𝒄𝒚 = 1/(2𝑥^2 − 1) 𝟏/𝐜𝐨𝐬𝒚 = 1/(2𝑥^2 − 1) cos𝑦 = 2𝑥2−1 y = cos –1 (2𝑥2−1) Putting 𝑥 = cosθ 𝑦 = cos –1 (2𝑐𝑜𝑠2𝜃−1) 𝑦 = cos –1 (cos2 𝜃) 𝑦 = 2𝜃 Putting value of θ = cos−1 x 𝑦 = 2 〖𝑐𝑜𝑠〗^(−1) 𝑥 Differentiating both sides 𝑤.𝑟.𝑡.𝑥 . (𝑑(𝑦))/𝑑𝑥 = (𝑑 (2〖𝑐𝑜𝑠〗^(−1) 𝑥" " ))/𝑑𝑥 𝑑𝑦/𝑑𝑥 = 2 (𝑑 (〖𝑐𝑜𝑠〗^(−1) 𝑥" " ))/𝑑𝑥 𝑑𝑦/𝑑𝑥 = 2 . ((−1)/√(1 − 𝑥^2 )) 𝒅𝒚/𝒅𝒙 = (−𝟐)/√(𝟏 − 𝒙^𝟐 ) (𝑐𝑜𝑠2𝜃 " = 2 " 〖𝑐𝑜𝑠〗^2 𝜃−1) Since x = cos θ ∴ 〖𝑐𝑜𝑠〗^(−1) x = θ ((〖𝑐𝑜𝑠〗^(−1) 𝑥")‘ = " (−1)/√(1 − 𝑥^2 ))
