MHB What Is the Derivative of \( y = x \sin x \)?

[karush]

v=197
If $y=x \sin x,$ then $\dfrac{dy}{dx}=$

$a.\quad\sin{x}+\cos{x}$
$b.\quad\sin{x}+x\cos{x}$
$c.\quad\sin{x}+\cos{x}$
$d.\quad x(\sin{x}+\cos{x})$
$e.\quad x(\sin{x}-\cos{x})$
well just by looking at it because $dx(x) = 1$
elimanates all the options besides b

$1\cdot \sin (x)+\cos (x)x$ or $\sin (x)+x\cos (x)$
otherwise the gymnastics of the product rule
$uv'+u'v$

 

[topsquark]

Yup. You got it.

-Dan