differentiation(3)

a) u = (2 - 1 - 2x)/(1 + 2x)

= 2/(1 + 2x) - 1

hence du/dx = -4/(1 + 2x)2.

b) y2 = sec u

Taking diff.:

2y dy/du = sec u tan u

dy/du = (tan u √sec u)/2

dy/dx = (dy/du)(du/dx)

= [(tan u √sec u)/2] [-4/(1 + 2x)2]

From (a), u = 2/(1 + 2x) - 1

u + 1 = 2/(1 + 2x)

(u + 1)2 = 4/(1 + 2x)2

So:

dy/dx = [(tan u √sec u)/2] [-4/(1 + 2x)2]

= - (u + 1)2 (tan u √sec u)/2