Check sibling questions

If y = Determinant | f(x) g(x) h(x) l m n a b c|, prove dy/dx = | f'(x

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

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


Transcript

Misc 22 (Method 1) If 𝑦 = |█( 𝑓(𝑥) 𝑔(𝑥) ℎ(𝑥)@𝑙 𝑚 𝑛@𝑎 𝑏 Here 𝑑𝑦/𝑑𝑥 = |█( 𝑓′(𝑥) 𝑔′(𝑥) ℎ′(𝑥)@𝑙 𝑚 𝑛@𝑎 𝑏 𝑐 )| Expanding determinant 𝑑𝑦/𝑑𝑥 = |𝑓′(𝑥)| |■8(𝑚&𝑛@𝑏&𝑐)||−𝑔′(𝑥) | |■8(𝑙&𝑛@𝑎&𝑐)||1+ ℎ′(𝑥) ||■8(𝑙&𝑚@𝑎&𝑏)| 𝑑𝑦/𝑑𝑥 = 𝑓′(𝑥) (𝑚𝑐 −𝑏𝑛)−𝑔′(𝑛) (𝑙𝑐−𝑎𝑛) + ℎ′(𝑛) (𝑙𝑏−𝑎𝑚) 𝑑𝑦/𝑑𝑥 = (𝑚𝑐 −𝑏𝑛) 𝑓′(𝑥)−(𝑙𝑐−𝑎𝑛)𝑔′(𝑥) +(𝑙𝑏−𝑎𝑚) ℎ′(𝑥) Hence We need to prove that 𝒅𝒚/𝒅𝒙 = (𝑚𝑐 −𝑏𝑛) 𝑓′(𝑥)−(𝑙𝑐−𝑎𝑛)𝑔′(𝑥) +(𝑙𝑏−𝑎𝑚) ℎ′(𝑥) Now, 𝑦 = |█( 𝑓(𝑥) 𝑔(𝑥) ℎ(𝑥)@𝑙 𝑚 𝑛@𝑎 𝑏 𝑐 )| Expanding determinant 𝑦 = 𝑓(𝑥)|■8(𝑚&𝑛@𝑏&𝑐)|− 𝑔(𝑥)|■8(𝑙&𝑛@𝑎&𝑐)|+ ℎ(𝑥)|■8(𝑙&𝑚@𝑎&𝑏)| 𝑦 = 𝑓(𝑥) (𝑚𝑐 −𝑏𝑛)−𝑔(𝑛) (𝑙𝑐−𝑎𝑛) + ℎ(𝑛) (𝑙𝑏−𝑎𝑚) 𝑦 = (𝑚𝑐 −𝑏𝑛) 𝑓(𝑥)−(𝑙𝑐−𝑎𝑛)𝑔(𝑥)" +" (𝑙𝑏−𝑎𝑚) ℎ(𝑥)" " Differentiating 𝑤.𝑟.𝑡.𝑥. 𝑑𝑦/𝑑𝑥 = 𝑑((𝑚𝑐 − 𝑏𝑛) 𝑓(𝑥) − (𝑙𝑐 − 𝑎𝑛)𝑔(𝑥)" +" (𝑙𝑏 − 𝑎𝑚) ℎ(𝑥)" " )/𝑑𝑥 𝑑𝑦/𝑑𝑥 = 𝑑((𝑚𝑐 − 𝑏𝑛) 𝑓(𝑥))/𝑑𝑥 − 𝑑((𝑙𝑐 − 𝑎𝑛)𝑔(𝑥))/𝑑𝑥 + 𝑑((𝑙𝑏 − 𝑎𝑚) ℎ(𝑥))/𝑑𝑥 𝑑𝑦/𝑑𝑥 = (𝑚𝑐−𝑏𝑛) 𝑑(𝑓(𝑥))/𝑑𝑥 − (𝑙𝑐−𝑎𝑛) 𝑑(𝑔(𝑥))/𝑑𝑥 + (𝑙𝑏−𝑎𝑚) 𝑑(ℎ(𝑥))/𝑑𝑥 𝑑𝑦/𝑑𝑥 = (𝑚𝑐−𝑏𝑛) 𝑓′(𝑥)−(𝑙𝑐−𝑎𝑛) 𝑔′(𝑥) + (𝑙𝑏−𝑎𝑚) ℎ′(𝑥)" " Hence proved Misc 22 (Method 2) If 𝑦 = |█( 𝑓(𝑥) 𝑔(𝑥) ℎ(𝑥)@𝑙 𝑚 𝑛@𝑎 𝑏 𝑐 )| , prove that 𝑑𝑦/𝑑𝑥 = |█( 𝑓′(𝑥) 𝑔′(𝑥) ℎ′(𝑥)@𝑙 𝑚 𝑛@𝑎 𝑏 𝑐 )| To Differentiate a determinant, We differentiate one row (or one column) at a time keeping others unchanged If 𝑦 = |█( 𝑓(𝑥) 𝑔(𝑥) ℎ(𝑥)@𝑙 𝑚 𝑛@𝑎 𝑏 𝑐 )| 𝑑𝑦/𝑑𝑥 = |█( 𝑓′(𝑥) 𝑔′(𝑥) ℎ′(𝑥)@𝑙 𝑚 𝑛@𝑎 𝑏 𝑐 )| + |█(𝑓(𝑥) 𝑔(𝑥) ℎ(𝑥)@(𝑙)^′ (𝑚)^′ (𝑛)^′@𝑎 𝑏 𝑐 )| + |█( 𝑓(𝑥) 𝑔(𝑥) ℎ(𝑥)@𝑙 𝑚 𝑛@(𝑎)′ (𝑏)′ (𝑐)′ )| 𝑑𝑦/𝑑𝑥 = |█( 𝑓′(𝑥) 𝑔′(𝑥) ℎ′(𝑥)@𝑙 𝑚 𝑛@𝑎 𝑏 𝑐 )| + |█(𝑓(𝑥) 𝑔(𝑥) ℎ(𝑥)@0 0 0 @𝑎 𝑏 𝑐 )| + |█( 𝑓(𝑥) 𝑔(𝑥) ℎ(𝑥)@𝑙 𝑚 𝑛@0 0 0 )| 𝑑𝑦/𝑑𝑥 = |█( 𝑓′(𝑥) 𝑔′(𝑥) ℎ′(𝑥)@𝑙 𝑚 𝑛@𝑎 𝑏 𝑐 )| + 0 + 0 𝑑𝑦/𝑑𝑥 = |█( 𝑓′(𝑥) 𝑔′(𝑥) ℎ′(𝑥)@𝑙 𝑚 𝑛@𝑎 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/𝑑𝑥 = |█( 𝑓′(𝑥) 𝑔′(𝑥) ℎ′(𝑥)@𝑙 𝑚 𝑛@𝑎 𝑏 𝑐 )|

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.