# How do you differentiate f(x)= (2 x^2 + 7 x - 6 )/ (x sinx ) using the quotient rule?

Jan 24, 2018

$f ' \left(x\right) = \frac{2 \sin x \left({x}^{2} + 3\right) - x \cos x \left(2 {x}^{2} + 7 x - 6\right)}{{\left(x \sin x\right)}^{2}}$

#### Explanation:

$f \left(x\right) = \frac{2 {x}^{2} + 7 x - 6}{x \sin x}$

Quotient rule: $\frac{d}{\mathrm{dx}} \left(\frac{f}{g}\right) = \frac{g {f}^{'} - f {g}^{'}}{g} ^ 2$

$\frac{d}{\mathrm{dx}} \left(2 {x}^{2} + 7 x - 6\right) = 4 x + 7$

$\frac{d}{\mathrm{dx}} \left(x \sin x\right) = \sin x + x \cos x$

$f ' \left(x\right) = \frac{x \sin x \left(4 x + 7\right) - \left(2 {x}^{2} + 7 x - 6\right) \left(\sin x + x \cos x\right)}{x \sin x} ^ 2$

$= \frac{4 {x}^{2} \sin x + \cancel{7 x \sin x} - 2 {x}^{2} \sin x - \cancel{7 x \sin x} + 6 \sin x - 2 {x}^{3} \cos x - 7 {x}^{2} \cos x + 6 x \cos x}{{\left(x \sin x\right)}^{2}}$

$= \frac{2 {x}^{2} \sin x + 6 \sin x - x \cos x \left(2 {x}^{2} + 7 x - 6\right)}{{\left(x \sin x\right)}^{2}}$

$= \frac{2 \sin x \left({x}^{2} + 3\right) - x \cos x \left(2 {x}^{2} + 7 x - 6\right)}{{\left(x \sin x\right)}^{2}}$

$\therefore f ' \left(x\right) = \frac{2 \sin x \left({x}^{2} + 3\right) - x \cos x \left(2 {x}^{2} + 7 x - 6\right)}{{\left(x \sin x\right)}^{2}}$ [Ans]