# How do you find the derivative of  (3+sin(x))/(3x+cos(x))?

May 20, 2016

Use quotient rule to obtain $\left(\frac{f}{g}\right) ' = \frac{3 x \cos x - 8}{9 {x}^{2} + 6 x \cos x + {\cos}^{2} x}$

#### Explanation:

Quotient rule: $\left(\frac{f}{g}\right) ' = \frac{f ' g - g ' f}{g} ^ 2$
Let $f = 3 + \sin x$; therefore $f ' = \cos x$
Let $g = 3 x + \cos x$; therefore $g ' = 3 - \sin x$
$\left(\frac{f}{g}\right) ' = \frac{f ' g - g ' f}{g} ^ 2$
$= \frac{\cos x \left(3 x + \cos x\right) - \left(3 - \sin x\right) \left(3 + \sin x\right)}{3 x + \cos x} ^ 2$
$= \frac{\left(3 x \cos x + {\cos}^{2} x\right) - \left(9 - {\sin}^{2} x\right)}{9 {x}^{2} + 6 x \cos x + {\cos}^{2} x}$
$= \frac{3 x \cos x + {\sin}^{2} x + {\cos}^{2} x - 9}{9 {x}^{2} + 6 x \cos x + {\cos}^{2} x}$
Identity: ${\sin}^{2} x + {\cos}^{2} x = 1 \implies \left(\frac{f}{g}\right) ' = \frac{3 x \cos x - 8}{9 {x}^{2} + 6 x \cos x + {\cos}^{2} x}$