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

Jan 29, 2016

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

#### Explanation:

The quotient rule states that

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

Here, we have

$g \left(x\right) = 2 {x}^{2} - x - 6$
$h \left(x\right) = x \sin x$

To differentiate $g \left(x\right)$, use the power rule on each term.

$g ' \left(x\right) = 4 x - 1$

To differentiate $h \left(x\right)$, use the product rule.

$h ' \left(x\right) = \sin x + x \cos x$

Plug these into the original quotient rule expression.

$f ' \left(x\right) = \frac{\left(4 x - 1\right) \left(x \sin x\right) - \left(\sin x + x \cos x\right) \left(2 {x}^{2} - x - 6\right)}{{x}^{2} {\sin}^{2} x}$

We can distribute and simplify, but the answer can't really get much prettier.

$f ' \left(x\right) = \frac{4 {x}^{2} \sin x - x \sin x - 2 {x}^{2} \sin x + x \sin x + 6 \sin x - 2 {x}^{3} \cos x + {x}^{2} \cos x + 6 x \cos x}{{x}^{2} {\sin}^{2} x}$

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