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

Question

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

Christopher Allers · Accepted Answer

By the rule of quotient:

#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#

Simplifying is all that's left to do.

#=>(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#

Charles Arocha · Answer

undefined To differentiate $ f(x) = \frac{6x^2 + 3x - 6}{x - 1} $ using the quotient rule, you apply the formula:

$$ \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{vu' - uv'}{v^2} $$

where $ u = 6x^2 + 3x - 6 $ and $ v = x - 1 $.

Now, differentiate $ u $ and $ v $ separately:

$$ u' = 12x + 3 $$
$$ v' = 1 $$

Then apply the quotient rule formula:

$$ f'(x) = \frac{(x - 1)(12x + 3) - (6x^2 + 3x - 6)(1)}{(x - 1)^2} $$

$$ f'(x) = \frac{12x^2 - 12x + 3x - 3 - 6x^2 - 3x + 6}{(x - 1)^2} $$

$$ f'(x) = \frac{6x^2 - 12x + 3}{(x - 1)^2} $$