Question

How do you differentiate `f(x)= ( x + 1 )/ ( x - 6)` using the quotient rule?

Answer 1

`f' (x)=(-7)/(x-6)^2`

Explanation:

Use the formula `d/dx(u/v)=(v*d/dx(u)-u*d/dx(v))/v^2`

Given `f(x)=(x+1)/(x-6)`

Let `u=(x+1)` and `v=(x-6)`

By the formula

`d/dx(u/v)=(v*d/dx(u)-u*d/dx(v))/v^2`

`d/dx((x+1)/(x-6))=((x-6)*d/dx(x+1)-(x+1)*d/dx(x-6))/(x-6)^2`

`d/dx((x+1)/(x-6))=((x-6)(1+0)-(x+1)(1-0))/(x-6)^2`

`d/dx((x+1)/(x-6))=((x-6)-(x+1))/(x-6)^2=(x-6-x-1)/(x-6)^2`

`d/dx((x+1)/(x-6))=(-7)/(x-6)^2`

Second solution #2, Just to check the above solution:

Simplify the given first so that

`f(x)=(x+1)/(x-6)=1+7/(x-6)=1+7(x-6)^-1`

differentiate

`f(x)=1+7(x-6)^-1`

`f' (x)=0+7*(-1)*(x-6)^(-1-1)*d/dx(x-6)`

`f' (x)=-7(x-6)^-2`

`f' (x)=(-7)/(x-6)^2`

God bless....I hope the explanation is useful.