How do you find the derivative of quotient $\frac{1 - 3 x}{1 + 3 x}$ $(1-3x)/(1+3x)$ ?

Question

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Answer 1 · 2015-04-12T17:44:52

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 · 2015-04-12T18:09:25

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

Detail

Let

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

Differentiating both sides with respect to 'x'

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

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

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

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

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

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

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

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

How do you find the derivative of quotient (1-3x)/(1+3x)1−3x1+3x(1-3x)/(1+3x)?