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/𝑑𝑥, 𝑥 =(〖𝑠𝑖𝑛〗^3 𝑡)/√(cos⁡2𝑡 ) , 𝑦 = (〖𝑐𝑜𝑠〗^3 𝑡)/√(cos⁡2𝑡 )Here, 𝑑𝑦/𝑑𝑥 = (𝑑𝑦/𝑑𝑡)/(𝑑𝑥/𝑑𝑡) Calculating 𝒅𝒚/𝒅𝒕 𝑦 = (〖𝑐𝑜𝑠〗^3 𝑡)/√(cos⁡2𝑡 ) 𝑑𝑦/𝑑𝑡 " " = 𝑑/𝑑𝑡 ((〖𝑐𝑜𝑠〗^3 𝑡)/√(cos⁡2𝑡 )) 𝑑𝑦/𝑑𝑡 " " = (𝑑(〖𝑐𝑜𝑠〗^3 𝑡)/𝑑𝑡 . √(cos⁡2 𝑡) − 𝑑(√(cos⁡2𝑡 ))/𝑑𝑡 .〖 cos^3〗⁡𝑡)/(√(cos⁡2 𝑡))^2 𝑑𝑦/𝑑𝑡 " = " (3 cos^2⁡〖𝑡 〗. 𝑑(cos⁡𝑡 )/𝑑𝑡. √(cos⁡2 𝑡) − 1/(2√(cos⁡2𝑡 )) . 𝑑(cos⁡2𝑡 )/𝑑𝑡 .〖 cos^3〗⁡𝑡)/(√(cos⁡2 𝑡))^2 Using quotient rule As (𝑢/𝑣)^′ = (𝑢^′ 𝑣 − 𝑣^′ 𝑢)/𝑣^2 𝑑𝑦/𝑑𝑡 " = " (3 cos^2⁡〖𝑡 〗. (−sin⁡𝑡 ) . √(cos⁡2 𝑡) − 1/(2√(cos⁡2𝑡 )) . (−2 sin⁡2𝑡) .〖 cos^3〗⁡𝑡)/(√(cos⁡2 𝑡))^2 𝑑𝑦/𝑑𝑡 " =" (−3 cos^2⁡〖𝑡 〗 sin⁡𝑡 √(cos⁡2 𝑡) + 1/√(cos⁡2𝑡 ) . sin⁡2𝑡 .〖 cos^3〗⁡𝑡)/(√(cos⁡2 𝑡))^2 𝑑𝑦/𝑑𝑡 " =" ((−3 cos^2⁡〖𝑡 〗 sin⁡𝑡 √(cos⁡2 𝑡) × √(cos⁡2𝑡 ) + sin⁡2𝑡 .〖 cos^3〗⁡𝑡)/√(cos⁡2𝑡 ))/(√(cos⁡2 𝑡))^2 𝑑𝑦/𝑑𝑡 " =" (−3 cos^2⁡〖𝑡 〗 sin⁡𝑡 (cos⁡2 𝑡) + sin⁡2𝑡 .〖 cos^3〗⁡𝑡)/((√(cos⁡2 𝑡))^2 (√(cos⁡2 𝑡)) ) 𝑑𝑦/𝑑𝑡 " =" ( cos^2⁡𝑡 (−3 sin⁡𝑡 .cos⁡2𝑡 +〖 cos〗⁡𝑡 . sin⁡2𝑡 ))/((cos⁡2 𝑡)^(3/2) ) Calculating 𝒅𝒙/𝒅𝒕 𝑥 = (〖𝑠𝑖𝑛〗^3 𝑡)/√(cos⁡2𝑡 ) 𝑑𝑥/𝑑𝑡 = 𝑑/𝑑𝑥 ((〖𝑠𝑖𝑛〗^3 𝑡)/√(cos⁡2𝑡 )) 𝑑𝑥/𝑑𝑡 = (𝑑(〖𝑠𝑖𝑛〗^3 𝑡)/𝑑𝑡 . √(cos⁡2𝑡 ) − (𝑑(√(cos⁡2𝑡 )) )/𝑑𝑥 . 〖 𝑠𝑖𝑛〗^3 𝑡 )/(√(cos⁡2𝑡 ))^2 𝑑𝑥/𝑑𝑡 = (3 〖𝑠𝑖𝑛〗^2 𝑡 . (𝑑(sin⁡𝑡 ) )/𝑑𝑡 . √(cos⁡2𝑡 ) − 1/(2√(cos⁡2𝑡 )) . (𝑑(cos⁡2𝑡 ) )/𝑑𝑥 . 〖 𝑠𝑖𝑛〗^3 𝑡 )/(√(cos⁡2𝑡 ))^2 𝑑𝑥/𝑑𝑡 = (3 〖𝑠𝑖𝑛〗^2 𝑡 . cos⁡𝑡 . √(cos⁡〖2 𝑡〗 ) − 1/(2√(cos⁡〖2 𝑡〗 )) . (−sin⁡2𝑡 ) . 2 . 〖 𝑠𝑖𝑛〗^3 𝑡 )/((cos⁡〖2 𝑡〗 ) ) 𝑑𝑥/𝑑𝑡 = (3 〖𝑠𝑖𝑛〗^2 𝑡 . cos⁡𝑡 . (√(cos⁡2𝑡 )) . (√(cos⁡2𝑡 )) + sin⁡2𝑡 . 〖 𝑠𝑖𝑛〗^3 𝑡 )/((√(cos⁡2𝑡 )) (cos⁡2𝑡 ) ) 𝑑𝑥/𝑑𝑡 = (3 〖𝑠𝑖𝑛〗^2 𝑡 . cos⁡𝑡 . cos⁡2𝑡 + sin⁡2𝑡 . 〖 𝑠𝑖𝑛〗^3 𝑡 )/(cos⁡2𝑡 )^(3/2) 𝑑𝑥/𝑑𝑡 = (〖𝑠𝑖𝑛〗^2 𝑡 (3 cos⁡𝑡 . cos⁡2𝑡 + sin⁡2𝑡 . sin⁡𝑡 ) )/(cos⁡2𝑡 )^(3/2) Finding 𝒅𝒚/𝒅𝒙 𝒅𝒚/𝒅𝒙 = ((cos^2⁡𝑡 (−3 sin⁡𝑡 .cos⁡2𝑡 +〖 cos〗⁡𝑡 . sin⁡2𝑡 ))/((cos⁡2 𝑡)^(3/2) ))/((〖𝑠𝑖𝑛〗^2 𝑡 (3 cos⁡𝑡 . cos⁡2𝑡 + sin⁡2𝑡 . sin⁡𝑡 ) )/(cos⁡2𝑡 )^(3/2) ) 𝑑𝑦/𝑑𝑥 = (cos^2⁡𝑡 (−3 sin⁡𝑡 .cos⁡2𝑡 +〖 cos〗⁡𝑡 . sin⁡2𝑡 ))/(〖𝑠𝑖𝑛〗^2 𝑡 (3 cos⁡𝑡 . cos⁡2𝑡 + sin⁡2𝑡 . sin⁡𝑡 ) ) 𝑑𝑦/𝑑𝑥 = (cos^2⁡𝑡 (−3 sin⁡𝑡 .cos⁡2𝑡 +〖 cos〗⁡𝑡 . sin⁡2𝑡 ))/(〖𝑠𝑖𝑛〗^2 𝑡 (3 cos⁡𝑡 . cos⁡2𝑡 + sin⁡2𝑡 . sin⁡𝑡 ) ) 𝑑𝑦/𝑑𝑥 = cot^2 𝑡 ((−3 sin⁡𝑡 .cos⁡2𝑡 +〖 cos〗⁡𝑡 . sin⁡2𝑡)/(3 cos⁡𝑡 . cos⁡2𝑡 + sin⁡2𝑡 . sin⁡𝑡 )) Taking cos 2t common 𝑑𝑦/𝑑𝑥 = cot^2 𝑡 ((cos⁡2𝑡 (−3 sin⁡𝑡 + cos⁡𝑡 sin⁡2𝑡/cos⁡2𝑡 ))/(cos⁡2𝑡 (3 cos⁡〖𝑡 〗+sin⁡𝑡 . sin⁡2𝑡/cos⁡2𝑡 ) )) 𝑑𝑦/𝑑𝑥 = cot^2 𝑡 ((−3 sin⁡𝑡 + cos⁡𝑡 sin⁡2𝑡/cos⁡2𝑡 )/(3 cos⁡〖𝑡 〗+〖 sin〗⁡𝑡 . sin⁡2𝑡/cos⁡2𝑡 )) 𝑑𝑦/𝑑𝑥 = cot^2 𝑡 ((−3 sin⁡𝑡 + cos⁡𝑡 tan⁡2𝑡)/(3 cos⁡〖𝑡 〗+〖 sin〗⁡𝑡 . tan⁡2𝑡 )) Taking cos t common 𝑑𝑦/𝑑𝑥 = cot^2 𝑡 ((cos⁡𝑡 (−3 sin⁡𝑡/cos⁡𝑡 + tan⁡2𝑡))/(cos⁡𝑡 (3 + sin⁡𝑡/cos⁡𝑡 . tan⁡2𝑡 )) 𝑑𝑦/𝑑𝑥 = cot^2 𝑡 ((−3 tan⁡𝑡 + tan⁡2𝑡)/(3 +〖 tan〗⁡𝑡 . tan⁡2𝑡 )) 𝑑𝑦/𝑑𝑥 = cot^2 𝑡 ((tan⁡2𝑡 − 3 tan⁡𝑡)/(3 +〖 tan〗⁡𝑡 . tan⁡2𝑡 )) Using tan 2𝜃 = (2 𝑡𝑎𝑛⁡𝜃)/(1 〖𝑡𝑎𝑛〗^2⁡𝜃 ) 𝑑𝑦/𝑑𝑥 = cot^2 𝑡 (((2 tan⁡𝑡)/(1 − tan^2⁡𝑡 ) − 3 tan⁡𝑡)/(3 + (tan⁡𝑡 ) ((2 tan⁡𝑡)/(1 −tan^2⁡𝑡 )) )) 𝑑𝑦/𝑑𝑥 = cot^2 𝑡 (((2 tan⁡𝑡 − 3 tan⁡𝑡 (1 − tan^2⁡𝑡 ))/((1 − tan^2⁡𝑡)))/((3 (1− tan^2⁡𝑡 ) + tan⁡𝑡 (2 tan⁡𝑡 ))/((1 − tan^2⁡𝑡)))) 𝑑𝑦/𝑑𝑥 = cot^2 𝑡 ((2 tan⁡𝑡 −3 tan⁡𝑡 (1 − tan^2⁡𝑡 ))/(3 (1− tan^2⁡𝑡 ) + tan⁡𝑡 (2 tan⁡𝑡 ) )) 𝑑𝑦/𝑑𝑥 = cot^2 𝑡 ((2 tan⁡𝑡 −3 tan⁡𝑡 + 3 tan^3⁡𝑡)/(3 − 3 tan^2⁡𝑡 + 2 tan^2⁡𝑡 )) 𝑑𝑦/𝑑𝑥 = cot^2 𝑡 ((−tan⁡𝑡 + 3 tan^3⁡𝑡 ) )/((3 −tan^2⁡𝑡 ) ) 𝑑𝑦/𝑑𝑥 = cot^2 𝑡 (−(tan⁡𝑡 −3 tan^3⁡𝑡 ) )/((3 −tan^2⁡𝑡 ) ) 𝑑𝑦/𝑑𝑥 = 〖−cot〗^2 𝑡 ((tan⁡𝑡 −3 tan^3⁡𝑡 ) )/((3 −tan^2⁡𝑡 ) ) Multiplying cot2 t to numerator 𝑑𝑦/𝑑𝑥 = −((cot^2⁡𝑡 × tan⁡𝑡 − 3 cot^2⁡𝑡 tan^3⁡𝑡)/(3 −tan^2⁡𝑡 )) 𝑑𝑦/𝑑𝑥 = − ((1/tan^2⁡𝑡 × tan⁡𝑡 − 3 × 1/tan^2⁡𝑡 ×tan^3⁡𝑡)/(3 −tan^2⁡𝑡 )) 𝑑𝑦/𝑑𝑥 = − ((1/tan⁡𝑡 .− 3 tan⁡𝑡 )/(3 − tan^2⁡𝑡 )) 𝑑𝑦/𝑑𝑥 = − (((1 −3 tan⁡𝑡 (tan⁡〖𝑡)〗)/tan⁡𝑡 )/(3 − tan^2⁡𝑡 )) 𝑑𝑦/𝑑𝑥 = − ((1 − 3 tan^2⁡𝑡 )/(tan⁡𝑡 (3 − tan^2⁡𝑡 ) )) 𝑑𝑦/𝑑𝑥 = − ((1 − 3 tan^2⁡𝑡 )/(3 tan⁡𝑡 −tan^3⁡𝑡 )) 𝑑𝑦/𝑑𝑥 = (−1)/(((3 tan⁡𝑡 −〖 tan〗^3⁡𝑡)/(1 −3 tan^2⁡𝑡 )) ) 𝑑𝑦/𝑑𝑥 = − ((1 − 3 tan^2⁡𝑡 )/(3 tan⁡𝑡 −tan^3⁡𝑡 )) 𝑑𝑦/𝑑𝑥 = (−1)/(((3 tan⁡𝑡 −〖 tan〗^3⁡𝑡)/(1 −3 tan^2⁡𝑡 )) ) 𝐴𝑠 tan⁡3𝑥=(3 tan⁡𝑥 − tan^3⁡𝑥)/(1 − 3 tan^2⁡𝑥 ) 𝑑𝑦/𝑑𝑥 = (−1)/tan⁡3𝑡 𝒅𝒚/𝒅𝒙 = −𝒄𝒐𝒕⁡𝟑𝒕

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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.