Chapter 5 Class 12 Continuity and Differentiability

Class 12
Important Questions for exams Class 12

Transcript

Example 40 (Method 1) Differentiate the following π€.π.π‘. π₯. (i) cos^(β1) (sinβ‘π₯) Let π(π₯) = cos^(β1) (sinβ‘π₯) π(π₯) = cos^(β1) (γcos γβ‘(π/2 βπ₯) ) π(π) = π/π βπ Differentiating π€.π.π‘.π₯ πβ(π₯) = (π (π/2))/ππ₯ β (π(π₯))/ππ₯ πβ(π₯) = 0 β 1 πβ(π) = β 1(π΄π  γ π ππ π γβ‘γ=γπππ  γβ‘γ(π/2 βπ₯)γ γ ) ("As " (π(π₯))/ππ₯ " = 1 & " π/2 " is constant" ) Example 40 (Method 2) Differentiate the following π€.π.π‘. π₯. (i) cos^(β1) (sinβ‘π₯) Let π(π₯) = cos^(β1) (sinβ‘π₯) Differentiating π€.π.π‘.π₯ πβ²(π₯) = (β1)/β(1 β γ(sinβ‘π₯)γ^2 ) Γ (sinβ‘π₯ )^β² πβ²(π₯) = (β1)/β(1 β sin^2β‘π₯ ) Γcosβ‘π₯ πβ²(π₯) = (β1)/β(cos^2β‘π₯ ) Γcosβ‘π₯ πβ²(π₯) = (β1)/cosβ‘π₯ Γcosβ‘π₯ πβ(π) = β1