Test: Derivatives: Product And Quotient Rule - Commerce MCQ

y = (1 + 1/x)/(1 - 1/x)
y = (x+1)/(x-1)
dy/dx = [(x-1)-(x+1)]/(x-1)²
= -2/(x-1)²

Given, y = (sinx+cosx)/(sinx−cosx)
∴ dydx = [(sinx−cosx)(cosx−sinx)−(sinx+cosx)(cosx+sinx)]/(sinx−cosx)² 
[by quotient rule]
= [(−sinx−cosx)² − (sinx+cosx)²]/(sinx−cosx)²
− [(sinx−cosx)² − (sinx+cosx)²]/(sinx−cosx)²
= − [(sin²x + cos²x − 2sinxcosx + sin²x + cos²x + 2sinxcosx)]/(sinx−cosx)²
= −2/(sinx−cosx)²
= -2/(sin²x + cos²x - 2sinxcosx)
= - 2/(1 - sin2x)

using these two rulesdifferentiation of  f(x) = x³ + x² + 3 will be 3x² + 2x

w’(z) = [(2-z)dy/dz(3z+9)- (3z+9)dy/dz(2-x)]/(2-z)²
[(2-z)(3) - (3z+9)(-1)]/(2-z)²
(6-3z+3z+9)/(2-z)²
= 15/(2-z)²

y = (px²+qx+r)/(ax+b)
⇒ dy/dx = d/dx(px²+qx+r)/(ax+b)
= [(ax+b) d/dx(px²+qx+r) − (px²+qx+r) d/dx(ax+b)]/(ax+b)²
= [(ax+b)(2px+q) − (px²+qx+r).a]/(ax+b)²
= [(2apx² + aqx + 2bpx + bq) − (apx² + aqx + ar)]/(ax+b)²
= [apx² + 2bpx + bq − ar]/(ax+b)²