Let [tex]f(x)=x\sin x[/tex]. Then the derivative is given by the limit
[tex]f'(x)=\displaystyle\lim_{h\to0}\frac{f(x+h)-f(x)}h=\lim_{h\to0}\frac{(x+h)\sin(x+h)-x\sin x}h[/tex]
Expand the numerator:
[tex](x+h)\sin(x+h)-x\sin x=x(\sin x\cos h+\sin h\cos x)+h(\sin x\cos h+\sin h\cos x)-x\sin x[/tex]
This can be rewritten as
[tex]x\sin x(\cos h-1)+x\cos x\sin h+h(\sin x\cos h+\sin h\cos x)[/tex]
So you're left with the limit
[tex]\displaystyle\lim_{h\to0}\frac{x\sin x(\cos h-1)+x\cos x\sin h+h(\sin x\cos h+\sin h\cos x)}h[/tex]
which you can split up as a sum of factored limits
[tex]\displaystyle x\sin x\lim_{h\to0}\frac{\cos h-1}h+x\cos x\lim_{h\to0}\frac{\sin h}h+\lim_{h\to0}\frac{h(\sin x\cos h+\sin h\cos x)}h[/tex]
The first two limits use two special limits,
[tex]\displaystyle\lim_{h\to0}\frac{1-\cos h}h=0\quad\text{and}\quad\lim_{h\to0}\frac{\sin h}h=1[/tex]
while for the last, you can cancel out the factors of [tex]h[/tex] in the numerator and denominator. You're left with
[tex]\displaystyle 0+x\cos x+\lim_{h\to0}(\sin x\cos h+\sin h\cos x)[/tex]
You have [tex]\lim\limits_{h\to0}\cos h=1[/tex] and [tex]\lim\limits_{h\to0}\sin h=0[/tex], so the derivative is
[tex]f'(x)=x\cos x+\sin x[/tex]
