How do I differentiate this using quotient rule?

Question

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

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Answer 1 · 2018-02-04T20:23:12

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

#(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)#