What is the derivative of sin(x)/x?

Jun 14, 2016

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

Explanation:

To find the derivative of a function in the form $f \frac{x}{g} \left(x\right)$, use the quotient rule:

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

For the function $\sin \frac{x}{x}$, we see that:

$f \left(x\right) = \sin \left(x\right) \text{ "=>" } {f}^{'} \left(x\right) = \cos \left(x\right)$

$g \left(x\right) = x \text{ "=>" } {g}^{'} \left(x\right) = 1$

Plugging these into the quotient rule, we see that:

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

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

Jun 14, 2016

$\frac{x \cos x - \sin x}{x} ^ 2.$
Rule : $\frac{d}{\mathrm{dx}} \left(f \frac{x}{g} \left(x\right)\right) = \frac{g \left(x\right) \cdot f ' \left(x\right) - f \left(x\right) \cdot g ' \left(x\right)}{g \left(x\right)} ^ 2$
$\therefore \frac{d}{\mathrm{dx}} \left(\sin \frac{x}{x}\right) = \frac{x \cdot \left(\sin x\right) ' - \left(\sin x\right) \left(x\right) '}{x} ^ 2 = \frac{x \cos x - \sin x}{x} ^ 2.$