# How do you differentiate f(x)=x^3sqrt(x-2)sinx using the product rule?

Mar 8, 2018

$f ' \left(x\right) = 3 {x}^{2} \sqrt{x - 2} \sin x + \frac{{x}^{3} \sin x}{2 \sqrt{x - 2}} + {x}^{3} \sqrt{x - 2} \cos x$

#### Explanation:

If $f \left(x\right) = g \left(x\right) h \left(x\right) j \left(x\right)$, then $f ' \left(x\right) = g ' \left(x\right) h \left(x\right) j \left(x\right) + g \left(x\right) h ' \left(x\right) j \left(x\right) + g \left(x\right) h \left(x\right) j ' \left(x\right)$

$g \left(x\right) = {x}^{3}$
$g ' \left(x\right) = 3 {x}^{2}$

$h \left(x\right) = \sqrt{x - 2} = {\left(x - 2\right)}^{\frac{1}{2}}$
$h ' \left(x\right) = \frac{1}{2} \cdot {\left(x - 2\right)}^{- \frac{1}{2}} \cdot \frac{d}{\mathrm{dx}} \left[x - 2\right]$
$\textcolor{w h i t e}{h ' \left(x\right)} = {\left(x - 2\right)}^{- \frac{1}{2}} / 2 \cdot 1$
$\textcolor{w h i t e}{h ' \left(x\right)} = {\left(x - 2\right)}^{- \frac{1}{2}} / 2$
$\textcolor{w h i t e}{h ' \left(x\right)} = \frac{1}{2 \sqrt{x - 2}}$

$j \left(x\right) = \sin x$
$j ' \left(x\right) = \cos x$

$f ' \left(x\right) = 3 {x}^{2} \sqrt{x - 2} \sin x + {x}^{3} \frac{1}{2 \sqrt{x - 2}} \sin x + {x}^{3} \sqrt{x - 2} \cos x$

$f ' \left(x\right) = 3 {x}^{2} \sqrt{x - 2} \sin x + \frac{{x}^{3} \sin x}{2 \sqrt{x - 2}} + {x}^{3} \sqrt{x - 2} \cos x$