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

2 Answers
Bdub
Mar 3, 2016

#f=3+x, g=1-3x, f'=1,g'=-3#
#f'(x)=((gf'-fg')/g^2) =( (1-3x)-(-3(3+x)))/(1-3x)^2->(1-3x+9+3x)/(1-3x)^2=10/((1-3x)^2#

Explanation:

Separate the top and the bottom into f and g then find the derivatives of each of them and then put it into the quotient rule #(gf'+fg')/g^2#

Jim G.
Mar 3, 2016

#f'(x) = -10/(1 - 3x )^2 #

Explanation:

differentiate using the#color(blue)" Quotient rule " #

If f(x)#= g(x)/(h(x) )" then " f'(x) =( g(x).h'(x) - h(x).g'(x))/(h(x))^2 #

here : g(x) = 3+x# " and " g'(x) = 1 #

and # h(x) = 1 - 3x" and " h'(x) = -3 #

substituting these results into f'(x)

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

#= (-9-3x-1+3x)/(1-3x)^2 = -10/(1-3x)^2#

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