Derivatives in parametric form

Chapter 5 Class 12 Continuity and Differentiability (Term 1)
Concept wise

### Transcript

Ex 5.6, 7 If x and y are connected parametrically by the equations without eliminating the parameter, Find ππ¦/ππ₯, π₯ =(γπ ππγ^3 π‘)/β(cosβ‘2π‘ ) , π¦ = (γπππ γ^3 π‘)/β(cosβ‘2π‘ )Here, ππ¦/ππ₯ = (ππ¦/ππ‘)/(ππ₯/ππ‘) Calculating ππ/ππ π¦ = (γπππ γ^3 π‘)/β(cosβ‘2π‘ ) ππ¦/ππ‘ " " = π/ππ‘ ((γπππ γ^3 π‘)/β(cosβ‘2π‘ )) ππ¦/ππ‘ " " = (π(γπππ γ^3 π‘)/ππ‘ . β(cosβ‘2 π‘) β π(β(cosβ‘2π‘ ))/ππ‘ .γ cos^3γβ‘π‘)/(β(cosβ‘2 π‘))^2 ππ¦/ππ‘ " = " (3 cos^2β‘γπ‘ γ. π(cosβ‘π‘ )/ππ‘. β(cosβ‘2 π‘) β 1/(2β(cosβ‘2π‘ )) . π(cosβ‘2π‘ )/ππ‘ .γ cos^3γβ‘π‘)/(β(cosβ‘2 π‘))^2 Using quotient rule As (π’/π£)^β² = (π’^β² π£ β π£^β² π’)/π£^2 ππ¦/ππ‘ " = " (3 cos^2β‘γπ‘ γ. (βsinβ‘π‘ ) . β(cosβ‘2 π‘) β 1/(2β(cosβ‘2π‘ )) . (β2 sinβ‘2π‘) .γ cos^3γβ‘π‘)/(β(cosβ‘2 π‘))^2 ππ¦/ππ‘ " =" (β3 cos^2β‘γπ‘ γ sinβ‘π‘ β(cosβ‘2 π‘) + 1/β(cosβ‘2π‘ ) . sinβ‘2π‘ .γ cos^3γβ‘π‘)/(β(cosβ‘2 π‘))^2 ππ¦/ππ‘ " =" ((β3 cos^2β‘γπ‘ γ sinβ‘π‘ β(cosβ‘2 π‘) Γ β(cosβ‘2π‘ ) + sinβ‘2π‘ .γ cos^3γβ‘π‘)/β(cosβ‘2π‘ ))/(β(cosβ‘2 π‘))^2 ππ¦/ππ‘ " =" (β3 cos^2β‘γπ‘ γ sinβ‘π‘ (cosβ‘2 π‘) + sinβ‘2π‘ .γ cos^3γβ‘π‘)/((β(cosβ‘2 π‘))^2 (β(cosβ‘2 π‘)) ) ππ¦/ππ‘ " =" ( cos^2β‘π‘ (β3 sinβ‘π‘ .cosβ‘2π‘ +γ cosγβ‘π‘ . sinβ‘2π‘ ))/((cosβ‘2 π‘)^(3/2) ) Calculating ππ/ππ π₯ = (γπ ππγ^3 π‘)/β(cosβ‘2π‘ ) ππ₯/ππ‘ = π/ππ₯ ((γπ ππγ^3 π‘)/β(cosβ‘2π‘ )) ππ₯/ππ‘ = (π(γπ ππγ^3 π‘)/ππ‘ . β(cosβ‘2π‘ ) β (π(β(cosβ‘2π‘ )) )/ππ₯ . γ π ππγ^3 π‘ )/(β(cosβ‘2π‘ ))^2 ππ₯/ππ‘ = (3 γπ ππγ^2 π‘ . (π(sinβ‘π‘ ) )/ππ‘ . β(cosβ‘2π‘ ) β 1/(2β(cosβ‘2π‘ )) . (π(cosβ‘2π‘ ) )/ππ₯ . γ π ππγ^3 π‘ )/(β(cosβ‘2π‘ ))^2 ππ₯/ππ‘ = (3 γπ ππγ^2 π‘ . cosβ‘π‘ . β(cosβ‘γ2 π‘γ ) β 1/(2β(cosβ‘γ2 π‘γ )) . (βsinβ‘2π‘ ) . 2 . γ π ππγ^3 π‘ )/((cosβ‘γ2 π‘γ ) ) ππ₯/ππ‘ = (3 γπ ππγ^2 π‘ . cosβ‘π‘ . (β(cosβ‘2π‘ )) . (β(cosβ‘2π‘ )) + sinβ‘2π‘ . γ π ππγ^3 π‘ )/((β(cosβ‘2π‘ )) (cosβ‘2π‘ ) ) ππ₯/ππ‘ = (3 γπ ππγ^2 π‘ . cosβ‘π‘ . cosβ‘2π‘ + sinβ‘2π‘ . γ π ππγ^3 π‘ )/(cosβ‘2π‘ )^(3/2) ππ₯/ππ‘ = (γπ ππγ^2 π‘ (3 cosβ‘π‘ . cosβ‘2π‘ + sinβ‘2π‘ . sinβ‘π‘ ) )/(cosβ‘2π‘ )^(3/2) Finding ππ/ππ ππ/ππ = ((cos^2β‘π‘ (β3 sinβ‘π‘ .cosβ‘2π‘ +γ cosγβ‘π‘ . sinβ‘2π‘ ))/((cosβ‘2 π‘)^(3/2) ))/((γπ ππγ^2 π‘ (3 cosβ‘π‘ . cosβ‘2π‘ + sinβ‘2π‘ . sinβ‘π‘ ) )/(cosβ‘2π‘ )^(3/2) ) ππ¦/ππ₯ = (cos^2β‘π‘ (β3 sinβ‘π‘ .cosβ‘2π‘ +γ cosγβ‘π‘ . sinβ‘2π‘ ))/(γπ ππγ^2 π‘ (3 cosβ‘π‘ . cosβ‘2π‘ + sinβ‘2π‘ . sinβ‘π‘ ) ) ππ¦/ππ₯ = (cos^2β‘π‘ (β3 sinβ‘π‘ .cosβ‘2π‘ +γ cosγβ‘π‘ . sinβ‘2π‘ ))/(γπ ππγ^2 π‘ (3 cosβ‘π‘ . cosβ‘2π‘ + sinβ‘2π‘ . sinβ‘π‘ ) ) ππ¦/ππ₯ = cot^2 π‘ ((β3 sinβ‘π‘ .cosβ‘2π‘ +γ cosγβ‘π‘ . sinβ‘2π‘)/(3 cosβ‘π‘ . cosβ‘2π‘ + sinβ‘2π‘ . sinβ‘π‘ )) Taking cos 2t common ππ¦/ππ₯ = cot^2 π‘ ((cosβ‘2π‘ (β3 sinβ‘π‘ + cosβ‘π‘ sinβ‘2π‘/cosβ‘2π‘ ))/(cosβ‘2π‘ (3 cosβ‘γπ‘ γ+sinβ‘π‘ . sinβ‘2π‘/cosβ‘2π‘ ) )) ππ¦/ππ₯ = cot^2 π‘ ((β3 sinβ‘π‘ + cosβ‘π‘ sinβ‘2π‘/cosβ‘2π‘ )/(3 cosβ‘γπ‘ γ+γ sinγβ‘π‘ . sinβ‘2π‘/cosβ‘2π‘ )) ππ¦/ππ₯ = cot^2 π‘ ((β3 sinβ‘π‘ + cosβ‘π‘ tanβ‘2π‘)/(3 cosβ‘γπ‘ γ+γ sinγβ‘π‘ . tanβ‘2π‘ )) Taking cos t common ππ¦/ππ₯ = cot^2 π‘ ((cosβ‘π‘ (β3 sinβ‘π‘/cosβ‘π‘ + tanβ‘2π‘))/(cosβ‘π‘ (3 + sinβ‘π‘/cosβ‘π‘ . tanβ‘2π‘ )) ππ¦/ππ₯ = cot^2 π‘ ((β3 tanβ‘π‘ + tanβ‘2π‘)/(3 +γ tanγβ‘π‘ . tanβ‘2π‘ )) ππ¦/ππ₯ = cot^2 π‘ ((tanβ‘2π‘ β 3 tanβ‘π‘)/(3 +γ tanγβ‘π‘ . tanβ‘2π‘ )) Using tan 2π = (2 π‘ππβ‘π)/(1 γπ‘ππγ^2β‘π ) ππ¦/ππ₯ = cot^2 π‘ (((2 tanβ‘π‘)/(1 β tan^2β‘π‘ ) β 3 tanβ‘π‘)/(3 + (tanβ‘π‘ ) ((2 tanβ‘π‘)/(1 βtan^2β‘π‘ )) )) ππ¦/ππ₯ = cot^2 π‘ (((2 tanβ‘π‘ β 3 tanβ‘π‘ (1 β tan^2β‘π‘ ))/((1 β tan^2β‘π‘)))/((3 (1β tan^2β‘π‘ ) + tanβ‘π‘ (2 tanβ‘π‘ ))/((1 β tan^2β‘π‘)))) ππ¦/ππ₯ = cot^2 π‘ ((2 tanβ‘π‘ β3 tanβ‘π‘ (1 β tan^2β‘π‘ ))/(3 (1β tan^2β‘π‘ ) + tanβ‘π‘ (2 tanβ‘π‘ ) )) ππ¦/ππ₯ = cot^2 π‘ ((2 tanβ‘π‘ β3 tanβ‘π‘ + 3 tan^3β‘π‘)/(3 β 3 tan^2β‘π‘ + 2 tan^2β‘π‘ )) ππ¦/ππ₯ = cot^2 π‘ ((βtanβ‘π‘ + 3 tan^3β‘π‘ ) )/((3 βtan^2β‘π‘ ) ) ππ¦/ππ₯ = cot^2 π‘ (β(tanβ‘π‘ β3 tan^3β‘π‘ ) )/((3 βtan^2β‘π‘ ) ) ππ¦/ππ₯ = γβcotγ^2 π‘ ((tanβ‘π‘ β3 tan^3β‘π‘ ) )/((3 βtan^2β‘π‘ ) ) Multiplying cot2 t to numerator ππ¦/ππ₯ = β((cot^2β‘π‘ Γ tanβ‘π‘ β 3 cot^2β‘π‘ tan^3β‘π‘)/(3 βtan^2β‘π‘ )) ππ¦/ππ₯ = β ((1/tan^2β‘π‘ Γ tanβ‘π‘ β 3 Γ 1/tan^2β‘π‘ Γtan^3β‘π‘)/(3 βtan^2β‘π‘ )) ππ¦/ππ₯ = β ((1/tanβ‘π‘ .β 3 tanβ‘π‘ )/(3 β tan^2β‘π‘ )) ππ¦/ππ₯ = β (((1 β3 tanβ‘π‘ (tanβ‘γπ‘)γ)/tanβ‘π‘ )/(3 β tan^2β‘π‘ )) ππ¦/ππ₯ = β ((1 β 3 tan^2β‘π‘ )/(tanβ‘π‘ (3 β tan^2β‘π‘ ) )) ππ¦/ππ₯ = β ((1 β 3 tan^2β‘π‘ )/(3 tanβ‘π‘ βtan^3β‘π‘ )) ππ¦/ππ₯ = (β1)/(((3 tanβ‘π‘ βγ tanγ^3β‘π‘)/(1 β3 tan^2β‘π‘ )) ) ππ¦/ππ₯ = β ((1 β 3 tan^2β‘π‘ )/(3 tanβ‘π‘ βtan^3β‘π‘ )) ππ¦/ππ₯ = (β1)/(((3 tanβ‘π‘ βγ tanγ^3β‘π‘)/(1 β3 tan^2β‘π‘ )) ) π΄π  tanβ‘3π₯=(3 tanβ‘π₯ β tan^3β‘π₯)/(1 β 3 tan^2β‘π₯ ) ππ¦/ππ₯ = (β1)/tanβ‘3π‘ ππ/ππ = βπππβ‘ππ