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