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

Jan 22, 2016

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

#### Explanation:

The quotient rule states that

$f ' \left(x\right) = \frac{\left(x - 6\right) \textcolor{red}{\frac{d}{\mathrm{dx}} \left[x \sin x\right]} - x \sin x \textcolor{b l u e}{\frac{d}{\mathrm{dx}} \left[x - 6\right]}}{x - 6} ^ 2$

Now, find both the internal derivatives.

The first requires product rule:

d/dx[xsinx]=sinxd/dx[x]+xd/dx[sinx]=color(red)(sinx+xcosx

The second is very simple.

$\frac{d}{\mathrm{dx}} \left[x - 6\right] = \textcolor{b l u e}{1}$

Plug these back in.

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

Simplify.

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

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