Question

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

Answer 1

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)`

Answer 2

`(-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)`