Finding derivative of Inverse trigonometric functions

Chapter 5 Class 12 Continuity and Differentiability
Concept wise

This video is only available for Teachoo black users

Introducing your new favourite teacher - Teachoo Black, at only βΉ83 per month

### Transcript

Misc 5 Differentiate π€.π.π‘. π₯ the function, (γπππ γ^(β1 ) π₯/2)/β(2π₯+7 ) , β 2 < π₯ < 2 Let π¦= (γπππ γ^(β1 ) π₯/2)/β(2π₯+7 ) Differentiating both sides π€.π.π‘. π₯ ππ¦/ππ₯ = π/ππ₯ ((γπππ γ^(β1 ) π₯/2)/β(2π₯+7 )) Using Quotient rule As (π’/π£)^β² = (π’^β² π£ β π£^β² π’)/π£^2 where u = cosβ1 π₯/2 & v = β(2π₯+7) ππ¦/ππ₯ = (π(γπππ γ^(β1 ) π₯/2)/ππ₯ . β(2π₯ + 7 ) β π(β(2π₯ + 7 ))/ππ₯ . γπππ γ^(β1 ) π₯/2 )/(β(2π₯ + 7 ))^2 ππ¦/ππ₯ = ((β1)/β(1 β(π₯/2)^2 ) . π(π₯/2)/ππ₯ β(2π₯ + 7 ) β 1/(2β(2π₯ + 7)) . π(2π₯ + 7)/ππ₯ . γπππ γ^(β1 ) π₯/2 )/(β(2π₯ + 7 ))^2 ππ¦/ππ₯ = ((β1)/β(γ(4 β π₯)/4γ^2 ) . 1/2 β(2π₯ + 7 ) β 1/(2β(2π₯ + 7)) . (2 + 0) . γπππ γ^(β1 ) π₯/2 )/((2π₯ + 7) ) ππ¦/ππ₯ = ((β2)/β(γ4 β π₯γ^2 ) . 1/2 β(2π₯ + 7 ) β 1/(2β(2π₯ + 7)) . 2 . γπππ γ^(β1 ) π₯/2 )/((2π₯ + 7) ) ππ¦/ππ₯ = ((β β(ππ + π ))/( β(γπ β πγ^π )) β (. γπππγ^(βπ ) π/π)/β(ππ + π) )/((2π₯ + 7) ) ππ¦/ππ₯ = (β β(2π₯ + 7 ) ( β(2π₯ + 7 )) β γπππ γ^(β1 ) π₯/2 β(γ4 β π₯γ^2 ))/((2π₯ + 7) (β(2π₯ + 7 )) (β(γ4 β π₯γ^2 )) ) ππ¦/ππ₯ = γβ(β(2π₯ + 7 ))γ^2/((2π₯ + 7) ( β(2π₯ + 7 )) (β(γ4 β π₯γ^2 )) ) "+ " (γπππ γ^(β1 ) π₯/2 β(γ4 β π₯γ^2 ))/((2π₯ + 7) ( β(2π₯ + 7 )) (β(γ4 β π₯γ^2 )) ) ππ¦/ππ₯ = (β(2π₯ + 7))/((2π₯ + 7) β(2π₯ + 7 ) β(γ4 β π₯γ^2 )) "+ " (γπππ γ^(β1 ) π₯/2 )/((2π₯ + 7) β(2π₯ + 7 ) ) ππ¦/ππ₯ = (β1)/(β(2π₯ + 7 ) β(γ4 β π₯γ^2 )) "+ " (γπππ γ^(β1 ) π₯/2 )/((2π₯ + 7) (2π₯ + 7)^(1/2) ) ππ/ππ = (βπ)/(β(ππ + π) β(γπ β πγ^π ))β(γπππγ^(βπ ) π/π )/((ππ + π)^(π/π) )