Answer
$y'=\dfrac{mx\cos mx-\sin mx}{x^{2}}$
Work Step by Step
$y=\dfrac{\sin mx}{x}$
Start the differentiation process by using the quotient rule:
$y'=\dfrac{(x)(\sin mx)'-(x)'(\sin mx)}{x^{2}}=...$
Apply the chain rule to evaluate the indicated derivatives and simplify:
$...=\dfrac{(x)(\cos mx)(mx)'-\sin mx}{x^{2}}=...$
$...=\dfrac{mx\cos mx-\sin mx}{x^{2}}$
