Ex 5.3, 13 - Chapter 5 Class 12 Continuity and Differentiability
Last updated at August 17, 2026 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, 13 Find ππ¦/ππ₯ in, y = cosβ1 (2π₯/( 1+ π₯2 )) , β1 < x < 1 π¦ = cosβ1 (2π₯/( 1+ π₯2 )) Let π₯ = tanβ‘π π¦ = cosβ1 ((2 tanβ‘π)/( 1 + π‘ππ2π )) π¦ = cosβ1 (sin 2ΞΈ) π¦ ="cosβ1" (γcos γβ‘(π/2 β2π) ) π¦ = π/2 β 2π Putting value of ΞΈ = tanβ1 x π¦ = π/2 β 2 γπ‘ππγ^(β1) π₯ Since x = tan ΞΈ β΄ γπ‘ππγ^(β1) x = ΞΈ Differentiating both sides π€.π.π‘.π₯ (π(π¦))/ππ₯ = (π (" " π/2 " β " γ2π‘ππγ^(β1) π₯" " ))/ππ₯ ππ¦/ππ₯ = 0 β 2 (πγ (π‘ππγ^(β1) π₯))/ππ₯ ππ¦/ππ₯ = β 2 (πγ (π‘ππγ^(β1) π₯))/ππ₯ ππ¦/ππ₯ = β 2 (1/(1 + π₯^2 )) π π/π π = (βπ)/(π + π^π ) ((γπ‘ππγ^(β1) π₯") β = " 1/(1 + π₯^2 ))
