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Misc  22 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Misc  22 - Chapter 5 Class 12 Continuity and Differentiability - Part 3

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Misc  22 - Chapter 5 Class 12 Continuity and Differentiability - Part 4

Misc  22 - Chapter 5 Class 12 Continuity and Differentiability - Part 5

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Misc 22 (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 22 (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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