Question

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

Answer 1

Using the quotient rule we have that

#f'(x)=((x+3)'*(x-3)-(x-3)'*(x+3))/(x-3)^2=> f'(x)=((x-3)-(x+3))/(x-3)^2=> f'(x)=-6/(x-3)^2#

Answer 2

`-6/(x - 3 )^2`

Explanation:

differentiate using the`color(blue)(" quotient rule ")`

for a rational function `f(x) = g(x)/(h(x))`

then: `f'(x) = (h(x).g'(x) - g(x).h'(x))/[h(x) ]^2`

applying this to the above function gives :

`d/dx(( x+3)/(x-3)) =( (x-3) d/dx(x+3) - (x+3) d/dx (x-3))/(x-3)^2`

`=( (x-3).1 - (x+3).1)/(x-3)^2 =( x-3 - x - 3)/(x-3)^2`

`= -6/(x-3)^2`