Form 4 Derivatives

For convenience sake, I don't write lim@x → 0 to be lim, where @x is the small increment of x.

1. y = (sinx)^2

dydx = lim [(y + @y) - y] / @x

= lim [(sin^2(x + @x) - sin^2x] / @x

= lim [(sin(x + @x) + sinx)(sin(x + @x) - sinx)] / @x

= lim [2sin{[(x + @x) + x]/2}cos{[(x + @x) - x]/2} 2cos{[(x + @x) + x]/2}sin{[(x + @x) - x]/2}] / @x

= 4 lim {sin(2x + @x)/2 cos(@x/2) cos(2x + @x)/2 sin(@x/2) } / @x

= lim [2sin(2x + @x)/2 cos(2x + @x)/2][2sin(@x/2) cos(@x/2)] / @x

= lim [sin(2x + @x) sin(@x)] / @x

= sin(2x) {lim sin(@x) / @x} ... (This limit equals 1)

= sin(2x)


2. y = e^x sinx

dy/dx = lim [(y + @y) - y] / @x

= lim {e^(x + @x) sin(x + @x) - e^x sinx} / @x

= lim {e^(x + @x) [sinx cos@x + cosx sin@x] - e^x sinx} / @x

= lim {e^x sinx [e^(@x) cos@x - 1] + e^(x + @x) cosx sin@x} / @x

= {e^x sinx lim [e^(@x) cos@x - 1]/@x} + cosx lim e^(x + @x) sin@x / @x

= {e^x sinx lim [e^(@x) [cos@x - sin@x]]} + cosx lim e^(x + @x) lim sin@x / @x

(L' Hospital Rule)

Put @x = 0

= e^x sinx (1)(1 - 0) + cosx e^(x + 0) (1)

= e^x (sinx + cosx)