# How do you differentiate (xcosx)/ (sinx+x) using the quotient rule?

Oct 19, 2017

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

#### Explanation:

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

$f \left(x\right) = \frac{x \cos x}{\sin x + x}$

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

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

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

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

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