Ex 5.3, 9 - 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, 9 Find ππ¦/ππ₯ in, y = sin^(β1) (2π₯/( 1 + 2π₯2 )) π¦ = sin^(β1) (2π₯/( 1 + 2π₯2 )) Putting x = tan ΞΈ π¦ = sin^(β1) (2π₯/( 1 + π₯2 )) π¦ = sin^(β1) ((2 π‘ππβ‘π)/(1 + γπ‘ππγ^2 π)) π¦ = sin^(β1) ( sinβ‘2ΞΈ) π¦ = 2π Putting value of ΞΈ = γπ‘ππγ^(β1) π₯ π¦ = 2 γπ‘ππγ^(β1) π₯ ("Since " π ππβ‘2π" = " (2 π‘ππβ‘π)/(1 + γπ‘ππγ^2 π)) Since x = tan ΞΈ β΄ γπ‘ππγ^(β1) x = ΞΈ Differentiating both sides π€.π.π‘.π₯ . (π(π¦))/ππ₯ = (π (2 γπ‘ππγ^(β1) π₯" ) " )/ππ₯ ππ¦/ππ₯ = 2 (π (γπ‘ππγ^(β1) π₯" ) " )/ππ₯ ππ¦/ππ₯ = 2 (1/(1+ π₯^2 )) π π/π π = π/(π+γ πγ^π ) ((γπ‘ππγ^(β1) π₯")β = " 1/(1 + π₯^2 ))
