Question

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

Answer 1

`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`

Answer 2

`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`