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

Mar 8, 2017

By the quotient rule:

$\frac{d}{\mathrm{dx}} \left[g \frac{x}{h} \left(x\right)\right] = \frac{h \left(x\right) g ' \left(x\right) - g \left(x\right) h ' \left(x\right)}{h \left(x\right)} ^ 2$

We have:

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

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

$g ' \left(x\right) = \frac{d}{\mathrm{dx}} \left(6 {x}^{2} + 3 x - 6\right) = 12 x + 3$

So $g ' \left(x\right) h \left(x\right) = \left(12 x + 3\right) \left(x - 1\right)$.

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

$h ' \left(x\right) = \frac{d}{\mathrm{dx}} \left(x - 1\right) = 1$

So $g \left(x\right) h ' \left(x\right) = \left(6 {x}^{2} + 3 x - 6\right) \left(1\right) = 6 {x}^{2} + 3 x - 6$

We now have $\left(12 x + 3\right) \left(x - 1\right) - \left(6 {x}^{2} + 3 x - 6\right)$.

For the last part of the derivative, we just put everything we found above over $h {\left(x\right)}^{2}$, which is ${\left(x - 1\right)}^{2}$.

$\implies \frac{\left(12 x + 3\right) \left(x - 1\right) - \left(6 {x}^{2} + 3 x - 6\right)}{x - 1} ^ 2$

All that's left to do is simplify.

$\implies \frac{12 {x}^{2} + 3 x - 12 x - 3 - 6 {x}^{2} - 3 x + 6}{x - 1} ^ 2$

$\implies \frac{6 {x}^{2} - 12 x + 3}{x - 1} ^ 2$

$\implies \frac{3 \left(2 {x}^{2} - 4 x + 1\right)}{x - 1} ^ 2$

$\therefore f ' \left(x\right) = \frac{3 \left(2 {x}^{2} - 4 x + 1\right)}{x - 1} ^ 2$