How do I differentiate this using quotient rule?

y=(x^2)/(1-x^2)^(1/2)

1 Answer
Feb 4, 2018

-(x(x^2-2))/(1-x^2)^(3/2)

Explanation:

quotient rule:

(Deltay)/(Deltax) v/u = ((u(Deltay)/(Deltax)v)-(v(Deltay)/(Deltax)u))/v^2

(Deltay)/(Deltax) x^2/((1-x^2)^(1/2)) = ((x^2(Deltay)/(Deltax)(1-x^2)^(1/2))-((1-x^2)^(1/2)(Deltay)/(Deltax)x^2))/((1-x^2)^(1/2))^2

using the chain rule:

(Deltay)/(Deltax)(1-x^2)^(1/2) = 1/2(1-x^2)^(-1/2) * (Deltay)/(Deltax) (1-x^2)

(Deltay)/(Deltax) (1-x^2) = 0 -2x = -2x

1/2(1-x^2)^(-1/2) * -2x = (-2x)/(2sqrt(1-x^2))

= x/(sqrt(1-x^2))

(Deltay)/(Deltax) (x^2) = 2x

((1-x^2)^(1/2))^2 = 1-x^2

(Deltay)/(Deltax) x^2/((1-x^2)^(1/2)) = ((x^2x/(sqrt(1-x^2)))-(2xsqrt(1-x^2)))/(1-x^2)

(x^2x/(sqrt(1-x^2)))/(1-x^2) = (x^2x)/(sqrt(1-x^2)) * 1/(1-x^2)

= x^3/(1-x^2)^(3/2)

(2xsqrt(1-x^2))/(1-x^2) = (2x)/((1-x^2)^(1/2))

x^3/(1-x^2)^(3/2) - (2x)/(1-x^2)^(1/2) = x^3/(1-x^2)^(3/2) - ((2x)(1-x^2))/(1-x^2)^(3/2)

(2x)(1-x^2) = 2x - 2x^3

x^3/(1-x^2)^(3/2) - (2x-2x^3)/(1-x^2)^(3/2) = (2x - x^3)/(1-x^2)^(3/2)

or -(x(x^2-2))/(1-x^2)^(3/2)