How do you find the derivative of quotient #(1-3x)/(1+3x)#?

2 Answers
Apr 12, 2015

The Quotient Rule for Derivatives says
#(d(g(x))/(h(x)))/(dx) = (g'(x)*h(x) - g(x)*h'(x))/(h^2(x))#

So
#(d ((1-3x)/ (1+3x)))/(dx)#

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

#= (-6)/(9x^2+6x+1)#

Apr 12, 2015

#(-6)/((1+3x)^2)#

Detail

Let

#y=(1-3x)/(1+3x)#

Differentiating both sides with respect to 'x'

#(dy)/(dx)=d/dx((1-3x)/(1+3x))#

Using quotient rule

#(dy)/(dx)=((1+3x)d/dx(1-3x)-(1-3x)d/dx(1+3x))/((1+3x)^2)#

#(dy)/(dx)=((1+3x)(0-3)-(1-3x)(0+3))/((1+3x)^2)#

#(dy)/(dx)=((-3)(1+3x)-(3)(1-3x))/((1+3x)^2)#

#(dy)/(dx)=(-3(1)+(-3)(3x)-3(1)-3(-3x))/((1+3x)^2)#

#(dy)/(dx)=(-3-9x-3+9x)/((1+3x)^2)#

#(dy)/(dx)=(-3-3+cancel(9x)-cancel(9x))/((1+3x)^2)#

#(dy)/(dx)=(-6)/((1+3x)^2)#