0y' = Lt [f(x+∆x) - f(x)]/∆x
∆x => 0
Therefore if y = sin x
y' = Lt [sin (x + ∆x) - sin x] / ∆x
∆x=>0
y' = Lt [sin x cos ∆x+cos x sin ∆x - sin x]/∆x
∆x => 0
=>Lt [ sin x cos ∆x - sin x + cos x sin ∆x]/∆x
∆x => 0
=> Lt [ sin x (cos ∆x - 1) + cos x sin ∆x]/ ∆x
∆x => 0s
=>sin x Lt(cos ∆x-1)/∆x+cos xLt(sin ∆x/∆x)
∆x => 0 ∆x=> 0
=> sin x • 0 + cos x• 1
=> dy/dx (sin x) = 0 + cos x
=> dy/dx (sin x ) = cos x
[ Lt sin x/x = 1 ] & Lt (cos x - 1)/x = 1-1 = 0]
x => 0 x => 0
