# How do you differentiate f(x)=(cosx-x)/(sin3x-x^2) using the quotient rule?

Mar 5, 2017

$\frac{\left(- \sin x - 1\right) \left(\sin 3 x - {x}^{2}\right) - \left(3 \cos 3 x - 2 x\right) \left(\cos x - x\right)}{\sin 3 x - {x}^{2}} ^ 2$

#### Explanation:

The quotient rule states that

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

Here,

$f \left(x\right) = \cos x - x$

so

$f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left[\cos x\right] - \frac{d}{\mathrm{dx}} \left[x\right] = - \sin x - 1$

and

$g \left(x\right) = \sin 3 x - {x}^{2}$

so

$g ' \left(x\right) = \frac{d}{\mathrm{dx}} \left[\sin 3 x\right] - \frac{d}{\mathrm{dx}} \left[{x}^{2}\right] = 3 \cos 3 x - 2 x$

We can put these functions into the formula for the quotient rule:

$\frac{d}{\mathrm{dx}} f \frac{x}{g} \left(x\right) = \frac{\left(- \sin x - 1\right) \left(\sin 3 x - {x}^{2}\right) - \left(3 \cos 3 x - 2 x\right) \left(\cos x - x\right)}{\sin 3 x - {x}^{2}} ^ 2$

which you could expand out to give

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

though this is more complicated and unnecessary.