不難付加maths第4回...加問篇..=.=+

連點解咁變都唔知...有D難搞...最好溫下書...

d ( x ) / dx = 1

d ( kx) / dx = k*d ( x ) / dx = k*1 = k where k is a constant

so, d ( 2t ) / dt = 2

d (x^n) / dx = n*x^(n-1) for any real nos. n

so, d ( t^2 ) / dt = 2*t^(2-1) = 2t

after u find dy/dx , u sub.( 代入 ) t into dy/dx , it is the ans~

hope u understand , but u really really better revision ! add oil la !