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Ex 5.6, 7 - Find dy/dx, x = sin^3 t / root (cos 2t), y = cos^3 t /

Ex 5.6, 7 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Ex 5.6, 7 - Chapter 5 Class 12 Continuity and Differentiability - Part 3 Ex 5.6, 7 - Chapter 5 Class 12 Continuity and Differentiability - Part 4 Ex 5.6, 7 - Chapter 5 Class 12 Continuity and Differentiability - Part 5 Ex 5.6, 7 - Chapter 5 Class 12 Continuity and Differentiability - Part 6 Ex 5.6, 7 - Chapter 5 Class 12 Continuity and Differentiability - Part 7 Ex 5.6, 7 - Chapter 5 Class 12 Continuity and Differentiability - Part 8 Ex 5.6, 7 - Chapter 5 Class 12 Continuity and Differentiability - Part 9 Ex 5.6, 7 - Chapter 5 Class 12 Continuity and Differentiability - Part 10

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Ex 5.6, 7 If x and y are connected parametrically by the equations without eliminating the parameter, Find 𝑑𝑦/𝑑π‘₯, π‘₯ =(〖𝑠𝑖𝑛〗^3 𝑑)/√(cos⁑2𝑑 ) , 𝑦 = (γ€–π‘π‘œπ‘ γ€—^3 𝑑)/√(cos⁑2𝑑 )Here, 𝑑𝑦/𝑑π‘₯ = (𝑑𝑦/𝑑𝑑)/(𝑑π‘₯/𝑑𝑑) Calculating π’…π’š/𝒅𝒕 𝑦 = (γ€–π‘π‘œπ‘ γ€—^3 𝑑)/√(cos⁑2𝑑 ) 𝑑𝑦/𝑑𝑑 " " = 𝑑/𝑑𝑑 ((γ€–π‘π‘œπ‘ γ€—^3 𝑑)/√(cos⁑2𝑑 )) 𝑑𝑦/𝑑𝑑 " " = (𝑑(γ€–π‘π‘œπ‘ γ€—^3 𝑑)/𝑑𝑑 . √(cos⁑2 𝑑) βˆ’ 𝑑(√(cos⁑2𝑑 ))/𝑑𝑑 .γ€– cos^3〗⁑𝑑)/(√(cos⁑2 𝑑))^2 𝑑𝑦/𝑑𝑑 " = " (3 cos^2⁑〖𝑑 γ€—. 𝑑(cos⁑𝑑 )/𝑑𝑑. √(cos⁑2 𝑑) βˆ’ 1/(2√(cos⁑2𝑑 )) . 𝑑(cos⁑2𝑑 )/𝑑𝑑 .γ€– cos^3〗⁑𝑑)/(√(cos⁑2 𝑑))^2 Using quotient rule As (𝑒/𝑣)^β€² = (𝑒^β€² 𝑣 βˆ’ 𝑣^β€² 𝑒)/𝑣^2 𝑑𝑦/𝑑𝑑 " = " (3 cos^2⁑〖𝑑 γ€—. (βˆ’sin⁑𝑑 ) . √(cos⁑2 𝑑) βˆ’ 1/(2√(cos⁑2𝑑 )) . (βˆ’2 sin⁑2𝑑) .γ€– cos^3〗⁑𝑑)/(√(cos⁑2 𝑑))^2 𝑑𝑦/𝑑𝑑 " =" (βˆ’3 cos^2⁑〖𝑑 γ€— sin⁑𝑑 √(cos⁑2 𝑑) + 1/√(cos⁑2𝑑 ) . sin⁑2𝑑 .γ€– cos^3〗⁑𝑑)/(√(cos⁑2 𝑑))^2 𝑑𝑦/𝑑𝑑 " =" ((βˆ’3 cos^2⁑〖𝑑 γ€— sin⁑𝑑 √(cos⁑2 𝑑) Γ— √(cos⁑2𝑑 ) + sin⁑2𝑑 .γ€– cos^3〗⁑𝑑)/√(cos⁑2𝑑 ))/(√(cos⁑2 𝑑))^2 𝑑𝑦/𝑑𝑑 " =" (βˆ’3 cos^2⁑〖𝑑 γ€— sin⁑𝑑 (cos⁑2 𝑑) + sin⁑2𝑑 .γ€– cos^3〗⁑𝑑)/((√(cos⁑2 𝑑))^2 (√(cos⁑2 𝑑)) ) 𝑑𝑦/𝑑𝑑 " =" ( cos^2⁑𝑑 (βˆ’3 sin⁑𝑑 .cos⁑2𝑑 +γ€– cos〗⁑𝑑 . sin⁑2𝑑 ))/((cos⁑2 𝑑)^(3/2) ) Calculating 𝒅𝒙/𝒅𝒕 π‘₯ = (〖𝑠𝑖𝑛〗^3 𝑑)/√(cos⁑2𝑑 ) 𝑑π‘₯/𝑑𝑑 = 𝑑/𝑑π‘₯ ((〖𝑠𝑖𝑛〗^3 𝑑)/√(cos⁑2𝑑 )) 𝑑π‘₯/𝑑𝑑 = (𝑑(〖𝑠𝑖𝑛〗^3 𝑑)/𝑑𝑑 . √(cos⁑2𝑑 ) βˆ’ (𝑑(√(cos⁑2𝑑 )) )/𝑑π‘₯ . γ€– 𝑠𝑖𝑛〗^3 𝑑 )/(√(cos⁑2𝑑 ))^2 𝑑π‘₯/𝑑𝑑 = (3 〖𝑠𝑖𝑛〗^2 𝑑 . (𝑑(sin⁑𝑑 ) )/𝑑𝑑 . √(cos⁑2𝑑 ) βˆ’ 1/(2√(cos⁑2𝑑 )) . (𝑑(cos⁑2𝑑 ) )/𝑑π‘₯ . γ€– 𝑠𝑖𝑛〗^3 𝑑 )/(√(cos⁑2𝑑 ))^2 𝑑π‘₯/𝑑𝑑 = (3 〖𝑠𝑖𝑛〗^2 𝑑 . cos⁑𝑑 . √(cos⁑〖2 𝑑〗 ) βˆ’ 1/(2√(cos⁑〖2 𝑑〗 )) . (βˆ’sin⁑2𝑑 ) . 2 . γ€– 𝑠𝑖𝑛〗^3 𝑑 )/((cos⁑〖2 𝑑〗 ) ) 𝑑π‘₯/𝑑𝑑 = (3 〖𝑠𝑖𝑛〗^2 𝑑 . cos⁑𝑑 . (√(cos⁑2𝑑 )) . (√(cos⁑2𝑑 )) + sin⁑2𝑑 . γ€– 𝑠𝑖𝑛〗^3 𝑑 )/((√(cos⁑2𝑑 )) (cos⁑2𝑑 ) ) 𝑑π‘₯/𝑑𝑑 = (3 〖𝑠𝑖𝑛〗^2 𝑑 . cos⁑𝑑 . cos⁑2𝑑 + sin⁑2𝑑 . γ€– 𝑠𝑖𝑛〗^3 𝑑 )/(cos⁑2𝑑 )^(3/2) 𝑑π‘₯/𝑑𝑑 = (〖𝑠𝑖𝑛〗^2 𝑑 (3 cos⁑𝑑 . cos⁑2𝑑 + sin⁑2𝑑 . sin⁑𝑑 ) )/(cos⁑2𝑑 )^(3/2) Finding π’…π’š/𝒅𝒙 π’…π’š/𝒅𝒙 = ((cos^2⁑𝑑 (βˆ’3 sin⁑𝑑 .cos⁑2𝑑 +γ€– cos〗⁑𝑑 . sin⁑2𝑑 ))/((cos⁑2 𝑑)^(3/2) ))/((〖𝑠𝑖𝑛〗^2 𝑑 (3 cos⁑𝑑 . cos⁑2𝑑 + sin⁑2𝑑 . sin⁑𝑑 ) )/(cos⁑2𝑑 )^(3/2) ) 𝑑𝑦/𝑑π‘₯ = (cos^2⁑𝑑 (βˆ’3 sin⁑𝑑 .cos⁑2𝑑 +γ€– cos〗⁑𝑑 . sin⁑2𝑑 ))/(〖𝑠𝑖𝑛〗^2 𝑑 (3 cos⁑𝑑 . cos⁑2𝑑 + sin⁑2𝑑 . sin⁑𝑑 ) ) 𝑑𝑦/𝑑π‘₯ = (cos^2⁑𝑑 (βˆ’3 sin⁑𝑑 .cos⁑2𝑑 +γ€– cos〗⁑𝑑 . sin⁑2𝑑 ))/(〖𝑠𝑖𝑛〗^2 𝑑 (3 cos⁑𝑑 . cos⁑2𝑑 + sin⁑2𝑑 . sin⁑𝑑 ) ) 𝑑𝑦/𝑑π‘₯ = cot^2 𝑑 ((βˆ’3 sin⁑𝑑 .cos⁑2𝑑 +γ€– cos〗⁑𝑑 . sin⁑2𝑑)/(3 cos⁑𝑑 . cos⁑2𝑑 + sin⁑2𝑑 . sin⁑𝑑 )) Taking cos 2t common 𝑑𝑦/𝑑π‘₯ = cot^2 𝑑 ((cos⁑2𝑑 (βˆ’3 sin⁑𝑑 + cos⁑𝑑 sin⁑2𝑑/cos⁑2𝑑 ))/(cos⁑2𝑑 (3 cos⁑〖𝑑 γ€—+sin⁑𝑑 . sin⁑2𝑑/cos⁑2𝑑 ) )) 𝑑𝑦/𝑑π‘₯ = cot^2 𝑑 ((βˆ’3 sin⁑𝑑 + cos⁑𝑑 sin⁑2𝑑/cos⁑2𝑑 )/(3 cos⁑〖𝑑 γ€—+γ€– sin〗⁑𝑑 . sin⁑2𝑑/cos⁑2𝑑 )) 𝑑𝑦/𝑑π‘₯ = cot^2 𝑑 ((βˆ’3 sin⁑𝑑 + cos⁑𝑑 tan⁑2𝑑)/(3 cos⁑〖𝑑 γ€—+γ€– sin〗⁑𝑑 . tan⁑2𝑑 )) Taking cos t common 𝑑𝑦/𝑑π‘₯ = cot^2 𝑑 ((cos⁑𝑑 (βˆ’3 sin⁑𝑑/cos⁑𝑑 + tan⁑2𝑑))/(cos⁑𝑑 (3 + sin⁑𝑑/cos⁑𝑑 . tan⁑2𝑑 )) 𝑑𝑦/𝑑π‘₯ = cot^2 𝑑 ((βˆ’3 tan⁑𝑑 + tan⁑2𝑑)/(3 +γ€– tan〗⁑𝑑 . tan⁑2𝑑 )) 𝑑𝑦/𝑑π‘₯ = cot^2 𝑑 ((tan⁑2𝑑 βˆ’ 3 tan⁑𝑑)/(3 +γ€– tan〗⁑𝑑 . tan⁑2𝑑 )) Using tan 2πœƒ = (2 π‘‘π‘Žπ‘›β‘πœƒ)/(1 γ€–π‘‘π‘Žπ‘›γ€—^2β‘πœƒ ) 𝑑𝑦/𝑑π‘₯ = cot^2 𝑑 (((2 tan⁑𝑑)/(1 βˆ’ tan^2⁑𝑑 ) βˆ’ 3 tan⁑𝑑)/(3 + (tan⁑𝑑 ) ((2 tan⁑𝑑)/(1 βˆ’tan^2⁑𝑑 )) )) 𝑑𝑦/𝑑π‘₯ = cot^2 𝑑 (((2 tan⁑𝑑 βˆ’ 3 tan⁑𝑑 (1 βˆ’ tan^2⁑𝑑 ))/((1 βˆ’ tan^2⁑𝑑)))/((3 (1βˆ’ tan^2⁑𝑑 ) + tan⁑𝑑 (2 tan⁑𝑑 ))/((1 βˆ’ tan^2⁑𝑑)))) 𝑑𝑦/𝑑π‘₯ = cot^2 𝑑 ((2 tan⁑𝑑 βˆ’3 tan⁑𝑑 (1 βˆ’ tan^2⁑𝑑 ))/(3 (1βˆ’ tan^2⁑𝑑 ) + tan⁑𝑑 (2 tan⁑𝑑 ) )) 𝑑𝑦/𝑑π‘₯ = cot^2 𝑑 ((2 tan⁑𝑑 βˆ’3 tan⁑𝑑 + 3 tan^3⁑𝑑)/(3 βˆ’ 3 tan^2⁑𝑑 + 2 tan^2⁑𝑑 )) 𝑑𝑦/𝑑π‘₯ = cot^2 𝑑 ((βˆ’tan⁑𝑑 + 3 tan^3⁑𝑑 ) )/((3 βˆ’tan^2⁑𝑑 ) ) 𝑑𝑦/𝑑π‘₯ = cot^2 𝑑 (βˆ’(tan⁑𝑑 βˆ’3 tan^3⁑𝑑 ) )/((3 βˆ’tan^2⁑𝑑 ) ) 𝑑𝑦/𝑑π‘₯ = γ€–βˆ’cotγ€—^2 𝑑 ((tan⁑𝑑 βˆ’3 tan^3⁑𝑑 ) )/((3 βˆ’tan^2⁑𝑑 ) ) Multiplying cot2 t to numerator 𝑑𝑦/𝑑π‘₯ = βˆ’((cot^2⁑𝑑 Γ— tan⁑𝑑 βˆ’ 3 cot^2⁑𝑑 tan^3⁑𝑑)/(3 βˆ’tan^2⁑𝑑 )) 𝑑𝑦/𝑑π‘₯ = βˆ’ ((1/tan^2⁑𝑑 Γ— tan⁑𝑑 βˆ’ 3 Γ— 1/tan^2⁑𝑑 Γ—tan^3⁑𝑑)/(3 βˆ’tan^2⁑𝑑 )) 𝑑𝑦/𝑑π‘₯ = βˆ’ ((1/tan⁑𝑑 .βˆ’ 3 tan⁑𝑑 )/(3 βˆ’ tan^2⁑𝑑 )) 𝑑𝑦/𝑑π‘₯ = βˆ’ (((1 βˆ’3 tan⁑𝑑 (tan⁑〖𝑑)γ€—)/tan⁑𝑑 )/(3 βˆ’ tan^2⁑𝑑 )) 𝑑𝑦/𝑑π‘₯ = βˆ’ ((1 βˆ’ 3 tan^2⁑𝑑 )/(tan⁑𝑑 (3 βˆ’ tan^2⁑𝑑 ) )) 𝑑𝑦/𝑑π‘₯ = βˆ’ ((1 βˆ’ 3 tan^2⁑𝑑 )/(3 tan⁑𝑑 βˆ’tan^3⁑𝑑 )) 𝑑𝑦/𝑑π‘₯ = (βˆ’1)/(((3 tan⁑𝑑 βˆ’γ€– tanγ€—^3⁑𝑑)/(1 βˆ’3 tan^2⁑𝑑 )) ) 𝑑𝑦/𝑑π‘₯ = βˆ’ ((1 βˆ’ 3 tan^2⁑𝑑 )/(3 tan⁑𝑑 βˆ’tan^3⁑𝑑 )) 𝑑𝑦/𝑑π‘₯ = (βˆ’1)/(((3 tan⁑𝑑 βˆ’γ€– tanγ€—^3⁑𝑑)/(1 βˆ’3 tan^2⁑𝑑 )) ) 𝐴𝑠 tan⁑3π‘₯=(3 tan⁑π‘₯ βˆ’ tan^3⁑π‘₯)/(1 βˆ’ 3 tan^2⁑π‘₯ ) 𝑑𝑦/𝑑π‘₯ = (βˆ’1)/tan⁑3𝑑 π’…π’š/𝒅𝒙 = βˆ’π’„π’π’•β‘πŸ‘π’•

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.