How do you differentiate #f(x)= (6 x^2 + 3 x - 6 )/ (x- 1 )# using the quotient rule?

1 Answer
Morgan
Mar 8, 2017

By the quotient rule:

#d/dx[g(x)/h(x)]=(h(x)g'(x)-g(x)h'(x))/[h(x)]^2#

We have:

#f(x)=(6x^2+3x-6)/(x-1)#

In our case, #g(x)=6x^2+3x-6# and #h(x)=x-1#. As shown in the definition above, we can begin by taking the derivative of #g(x)# and multiplying this by #h(x)#, which we leave alone.

#g'(x)=d/dx(6x^2+3x-6)=12x+3#

So #g'(x)h(x)=(12x+3)(x-1)#.

Next, we take the derivative of #h(x)# and multiply the result by #g(x)#, which we leave alone. We'll subtract that from what we found above.

#h'(x)=d/dx(x-1)=1#

So #g(x)h'(x)=(6x^2+3x-6)(1)=6x^2+3x-6#

We now have #(12x+3)(x-1)-(6x^2+3x-6)#.

For the last part of the derivative, we just put everything we found above over #h(x)^2#, which is #(x-1)^2#.

#=>((12x+3)(x-1)-(6x^2+3x-6))/(x-1)^2#

All that's left to do is simplify.

#=>(12x^2+3x-12x-3-6x^2-3x+6)/(x-1)^2#

#=>(6x^2-12x+3)/(x-1)^2#

#=>(3(2x^2-4x+1))/(x-1)^2#

#:.f'(x)=(3(2x^2-4x+1))/(x-1)^2#

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