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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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Misc 21 (Method 1) If 𝑦 = |β–ˆ( 𝑓(π‘₯) 𝑔(π‘₯) β„Ž(π‘₯)@𝑙 π‘š 𝑛@π‘Ž 𝑏 Here 𝑑𝑦/𝑑π‘₯ = |β–ˆ( 𝑓′(π‘₯) 𝑔′(π‘₯) β„Žβ€²(π‘₯)@𝑙 π‘š 𝑛@π‘Ž 𝑏 𝑐 )| Expanding determinant 𝑑𝑦/𝑑π‘₯ = |𝑓′(π‘₯)| |β– 8(π‘š&𝑛@𝑏&𝑐)||βˆ’π‘”β€²(π‘₯) | |β– 8(𝑙&𝑛@π‘Ž&𝑐)||1+ β„Žβ€²(π‘₯) ||β– 8(𝑙&π‘š@π‘Ž&𝑏)| 𝑑𝑦/𝑑π‘₯ = 𝑓′(π‘₯) (π‘šπ‘ βˆ’π‘π‘›)βˆ’π‘”β€²(𝑛) (π‘™π‘βˆ’π‘Žπ‘›) + β„Žβ€²(𝑛) (π‘™π‘βˆ’π‘Žπ‘š) 𝑑𝑦/𝑑π‘₯ = (π‘šπ‘ βˆ’π‘π‘›) 𝑓′(π‘₯)βˆ’(π‘™π‘βˆ’π‘Žπ‘›)𝑔′(π‘₯) +(π‘™π‘βˆ’π‘Žπ‘š) β„Žβ€²(π‘₯) Hence We need to prove that π’…π’š/𝒅𝒙 = (π‘šπ‘ βˆ’π‘π‘›) 𝑓′(π‘₯)βˆ’(π‘™π‘βˆ’π‘Žπ‘›)𝑔′(π‘₯) +(π‘™π‘βˆ’π‘Žπ‘š) β„Žβ€²(π‘₯) Now, 𝑦 = |β–ˆ( 𝑓(π‘₯) 𝑔(π‘₯) β„Ž(π‘₯)@𝑙 π‘š 𝑛@π‘Ž 𝑏 𝑐 )| Expanding determinant 𝑦 = 𝑓(π‘₯)|β– 8(π‘š&𝑛@𝑏&𝑐)|βˆ’ 𝑔(π‘₯)|β– 8(𝑙&𝑛@π‘Ž&𝑐)|+ β„Ž(π‘₯)|β– 8(𝑙&π‘š@π‘Ž&𝑏)| 𝑦 = 𝑓(π‘₯) (π‘šπ‘ βˆ’π‘π‘›)βˆ’π‘”(𝑛) (π‘™π‘βˆ’π‘Žπ‘›) + β„Ž(𝑛) (π‘™π‘βˆ’π‘Žπ‘š) 𝑦 = (π‘šπ‘ βˆ’π‘π‘›) 𝑓(π‘₯)βˆ’(π‘™π‘βˆ’π‘Žπ‘›)𝑔(π‘₯)" +" (π‘™π‘βˆ’π‘Žπ‘š) β„Ž(π‘₯)" " Differentiating 𝑀.π‘Ÿ.𝑑.π‘₯. 𝑑𝑦/𝑑π‘₯ = 𝑑((π‘šπ‘ βˆ’ 𝑏𝑛) 𝑓(π‘₯) βˆ’ (𝑙𝑐 βˆ’ π‘Žπ‘›)𝑔(π‘₯)" +" (𝑙𝑏 βˆ’ π‘Žπ‘š) β„Ž(π‘₯)" " )/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ = 𝑑((π‘šπ‘ βˆ’ 𝑏𝑛) 𝑓(π‘₯))/𝑑π‘₯ βˆ’ 𝑑((𝑙𝑐 βˆ’ π‘Žπ‘›)𝑔(π‘₯))/𝑑π‘₯ + 𝑑((𝑙𝑏 βˆ’ π‘Žπ‘š) β„Ž(π‘₯))/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ = (π‘šπ‘βˆ’π‘π‘›) 𝑑(𝑓(π‘₯))/𝑑π‘₯ βˆ’ (π‘™π‘βˆ’π‘Žπ‘›) 𝑑(𝑔(π‘₯))/𝑑π‘₯ + (π‘™π‘βˆ’π‘Žπ‘š) 𝑑(β„Ž(π‘₯))/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ = (π‘šπ‘βˆ’π‘π‘›) 𝑓′(π‘₯)βˆ’(π‘™π‘βˆ’π‘Žπ‘›) 𝑔′(π‘₯) + (π‘™π‘βˆ’π‘Žπ‘š) β„Žβ€²(π‘₯)" " Hence proved Misc 21 (Method 2) If 𝑦 = |β–ˆ( 𝑓(π‘₯) 𝑔(π‘₯) β„Ž(π‘₯)@𝑙 π‘š 𝑛@π‘Ž 𝑏 𝑐 )| , prove that 𝑑𝑦/𝑑π‘₯ = |β–ˆ( 𝑓′(π‘₯) 𝑔′(π‘₯) β„Žβ€²(π‘₯)@𝑙 π‘š 𝑛@π‘Ž 𝑏 𝑐 )| To Differentiate a determinant, We differentiate one row (or one column) at a time keeping others unchanged If 𝑦 = |β–ˆ( 𝑓(π‘₯) 𝑔(π‘₯) β„Ž(π‘₯)@𝑙 π‘š 𝑛@π‘Ž 𝑏 𝑐 )| 𝑑𝑦/𝑑π‘₯ = |β–ˆ( 𝑓′(π‘₯) 𝑔′(π‘₯) β„Žβ€²(π‘₯)@𝑙 π‘š 𝑛@π‘Ž 𝑏 𝑐 )| + |β–ˆ(𝑓(π‘₯) 𝑔(π‘₯) β„Ž(π‘₯)@(𝑙)^β€² (π‘š)^β€² (𝑛)^β€²@π‘Ž 𝑏 𝑐 )| + |β–ˆ( 𝑓(π‘₯) 𝑔(π‘₯) β„Ž(π‘₯)@𝑙 π‘š 𝑛@(π‘Ž)β€² (𝑏)β€² (𝑐)β€² )| 𝑑𝑦/𝑑π‘₯ = |β–ˆ( 𝑓′(π‘₯) 𝑔′(π‘₯) β„Žβ€²(π‘₯)@𝑙 π‘š 𝑛@π‘Ž 𝑏 𝑐 )| + |β–ˆ(𝑓(π‘₯) 𝑔(π‘₯) β„Ž(π‘₯)@0 0 0 @π‘Ž 𝑏 𝑐 )| + |β–ˆ( 𝑓(π‘₯) 𝑔(π‘₯) β„Ž(π‘₯)@𝑙 π‘š 𝑛@0 0 0 )| 𝑑𝑦/𝑑π‘₯ = |β–ˆ( 𝑓′(π‘₯) 𝑔′(π‘₯) β„Žβ€²(π‘₯)@𝑙 π‘š 𝑛@π‘Ž 𝑏 𝑐 )| + 0 + 0 𝑑𝑦/𝑑π‘₯ = |β–ˆ( 𝑓′(π‘₯) 𝑔′(π‘₯) β„Žβ€²(π‘₯)@𝑙 π‘š 𝑛@π‘Ž 𝑏 𝑐 )| Hence proved. Using property If any one Row or column is 0 , then value of determinate is also 0 𝑐 )| , prove that 𝑑𝑦/𝑑π‘₯ = |β–ˆ( 𝑓′(π‘₯) 𝑔′(π‘₯) β„Žβ€²(π‘₯)@𝑙 π‘š 𝑛@π‘Ž 𝑏 𝑐 )|

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Davneet Singh

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.