Pre-requisite Matrix differentiation 5. Function is a vector and the variable is a vector x=[x,x2,,x],y=[y1(x)2(x)…n(x) In a similar way Oy,(x)ay2(x) aym( OX OX T Oy,(x)Oy,(x ay dx V,(xoV,(X nxm 2/3/2021 PATTERN RECOGNITION
Pre-requisite Matrix differentiation 5. Function is a vector and the variable is a vector In a similar way 2/3/2021 PATTERN RECOGNITION 16 1 2 1 2 , ,..., , ( ), ( ),..., ( ) T T n m x y x x x = = x x x y y y 1 2 1 1 1 1 2 2 2 2 1 2 ( ) ( ) ( ) , ,..., ( ) ( ) ( ) , ,..., ( ) ( ) ( ) , ,..., m T m m n n n n m y y y x x x y y y d x x x d y y y x x x = x x x x x x y x x x x
Pre-requisite Matrix differentiation 5. Function is a vector and the variable is a vector ,y=[(x)y2(x)…n(x) Example x y 2(x))) x 2|,y1(x)=x2 2y2 (x)=x2+3x 3 d,r dy2(x) dx dx dy(r)dy2(x 2x10 1 dx d: dx 02x dyer dy2(x d dx 2/3/2021 PATTERN RECOGNITION
Pre-requisite Matrix differentiation 5. Function is a vector and the variable is a vector Example 2/3/2021 PATTERN RECOGNITION 17 1 2 1 2 , ,..., , ( ), ( ),..., ( ) T T n m x y x x x = = x x x y y y 𝒚 = 𝑦1 (𝒙) 𝑦2 (𝒙) , 𝒙 = 𝑥1 𝑥2 𝑥3 ,𝑦1 𝒙 = 𝑥1 2 − 𝑥2 , 𝑦2 𝒙 = 𝑥3 2 + 3𝑥2 𝑑𝒚 𝑇 𝑑𝒙 = 𝑑𝑦1 (𝒙) 𝑑𝑥1 𝑑𝑦2 (𝒙) 𝑑𝑥1 𝑑𝑦1 (𝒙) 𝑑𝑥2 𝑑𝑦2 (𝒙) 𝑑𝑥2 𝑑𝑦1 (𝒙) 𝑑𝑥3 𝑑𝑦2 (𝒙) 𝑑𝑥3 = 2𝑥1 0 −1 3 0 2𝑥3