If I was to do this using first principles, I'd evaluate the derivatives of $\displaystyle \displaystyle \begin{align*} x \end{align*}$ and $\displaystyle \displaystyle \begin{align*} \sin{x} \end{align*}$ by first principles, then prove the product rule using first principles and apply it

$\displaystyle \displaystyle \begin{align*} f_1(x) &= x \\ f_1 ' (x) &= \lim_{h \to 0}\frac{f_1(x + h) - f_1(x)}{h} \\ &= \lim_{h \to 0}\frac{x + h - x}{h} \\ &= \lim_{h \to 0}\frac{h}{h} \\ &= \lim_{h \to 0} 1 \\ &= 1 \\ \\ f_2(x) &= \sin{(x)} \\ f_2 '(x) &= \lim_{h \to 0}\frac{f_2(x + h) - f_2(x)}{h} \\ &= \lim_{h \to 0}\frac{\sin{(x+h)} - \sin{(x)}}{h} \\ &= \lim_{h \to 0}\frac{\sin{(x)}\cos{(h)} + \cos{(x)}\sin{(h)} - \sin{(x)}}{h} \\ &= \sin{(x)}\lim_{h \to 0}\frac{\cos{(h)} - 1}{h} + \cos{(x)}\lim_{h \to 0}\frac{\sin{(h)}}{h} \\ &= \sin{(x)} \cdot 0 + \cos{(x)} \cdot 1 \\ &= \cos{(x)} \end{align*}$

Now as for the product rule...

$\displaystyle \displaystyle \begin{align*} \left[ u(x) v(x) \right]' &= \lim_{h \to 0}\frac{u(x + h)v(x + h) - u(x)v(x)}{h} \\ &= \lim_{h \to 0}\frac{u(x+h)v(x+h) - u(x +h)v(x) + u(x +h)v(x) - u(x)v(x)}{h} \\ &= \lim_{h \to 0}u(x+h) \cdot \lim_{h \to 0}\frac{v(x+h) - v(x)}{h} + v(x)\cdot \lim_{h \to 0}\frac{u(x+h) - u(x)}{h} \\ &= u(x) v'(x) + u'(x) v(x) \end{align*}$

So that means

$\displaystyle \displaystyle \begin{align*} \left[ x\sin{(x)} \right]' &= x \left[ \sin{(x)} \right]' + (x)' \sin{(x)} \\ &= x\cos{(x)} + \sin{(x)} \end{align*}$