# How do you differentiate f(x)=xsinsqrtx using the chain rule?

Mar 27, 2017

$\sin \sqrt{x} + \frac{\cos \sqrt{x} \cdot \sqrt{x}}{2}$

#### Explanation:

In order to differentiate $f \left(x\right) = x \sin \sqrt{x}$, we need to use the product rule

Product rule: $f \left(x\right) \cdot g \left(x\right) = f ' \left(x\right) g \left(x\right) + g ' \left(x\right) f \left(x\right)$

$f \left(x\right) \cdot g \left(x\right) = f ' \left(x\right) g \left(x\right) + g ' \left(x\right) f \left(x\right)$
$x \cdot \sin \sqrt{x} = \frac{d}{\mathrm{dx}} x \cdot \sin \sqrt{x} + \frac{d}{\mathrm{dx}} \sin \sqrt{x} \cdot x$

$x ' = 1$
$\sin \sqrt{x} ' = \cos \sqrt{x} \cdot \frac{d}{\mathrm{dx}} \sqrt{x}$

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

$\sin \sqrt{x} + \cos \sqrt{x} \cdot \frac{1}{2 \sqrt{x}} \cdot x$
$\sin \sqrt{x} + \frac{\cos \sqrt{x} \cdot \sqrt{x}}{2}$