Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Chapter 5 Class 12 Continuity and Differentiability
Ex 5.1, 13
Ex 5.1, 16
Ex 5.1, 18 Important
Ex 5.1, 28 Important
Ex 5.1, 30 Important
Ex 5.1, 34 Important
Ex 5.2, 5
Ex 5.2, 9 Important
Ex 5.2, 10 Important
Ex 5.3, 10 Important
Ex 5.3, 14
Example 29 Important
Example 30 Important
Ex 5.5,6 Important
Ex 5.5, 7 Important
Ex 5.5, 11 Important
Ex 5.5, 16 Important
Ex 5.6, 7 Important
Ex 5.6, 11 Important
Example 38 Important
Ex 5.7, 14 Important
Question 4 Important Deleted for CBSE Board 2024 Exams
Question 5 Important Deleted for CBSE Board 2024 Exams
Example 39 (i)
Example 40 (i) You are here
Example 42 Important
Misc 6 Important
Misc 15 Important
Misc 16 Important
Misc 22 Important
Chapter 5 Class 12 Continuity and Differentiability
Last updated at June 2, 2023 by Teachoo
Example 40 (Method 1) Differentiate the following π€.π.π‘. π₯. (i) cos^(β1) (sinβ‘π₯) Let π(π₯) = cos^(β1) (sinβ‘π₯) π(π₯) = cos^(β1) (γcos γβ‘(π/2 βπ₯) ) π(π) = π /π βπ Differentiating π€.π.π‘.π₯ πβ(π₯) = (π (π/2))/ππ₯ β (π(π₯))/ππ₯ πβ(π₯) = 0 β 1 πβ(π) = β 1(π΄π γ π ππ π γβ‘γ=γπππ γβ‘γ(π/2 βπ₯)γ γ ) ("As " (π(π₯))/ππ₯ " = 1 & " π/2 " is constant" ) Example 40 (Method 2) Differentiate the following π€.π.π‘. π₯. (i) cos^(β1) (sinβ‘π₯) Let π(π₯) = cos^(β1) (sinβ‘π₯) Differentiating π€.π.π‘.π₯ πβ²(π₯) = (β1)/β(1 β γ(sinβ‘π₯)γ^2 ) Γ (sinβ‘π₯ )^β² πβ²(π₯) = (β1)/β(1 β sin^2β‘π₯ ) Γcosβ‘π₯ πβ²(π₯) = (β1)/β(cos^2β‘π₯ ) Γcosβ‘π₯ πβ²(π₯) = (β1)/cosβ‘π₯ Γcosβ‘π₯ πβ(π) = β1