How do I find the 1st derivative $F(x) = x \, \sin x$ using the definition of a derivate? Here's what I've got so far:
$\frac{dF}{dx}=\lim\limits_{h \rightarrow 0}\frac{F(a+h) - F(a)}{h}$
$\frac{P(a+h) - P(a)}{h}$ = $\frac{(a+h)(sin(a+h) - asin(a)}{h}$
= $\frac{a(\sin(a+h)) + h(\sin(a+h)) - a\sin(a)}{h}$
= $\frac{a(\sin(a+h) - \sin(a)) + h(\sin(a+h))}{h}$
= $\frac{a(2\cos(\frac{2a+h}{2}) - \sin(\frac{h}{2}))}{h} + \frac{h(\sin(a+h))}{h}$
= $\frac{a(2\cos(\frac{2a+h}{2}) - \sin(\frac{h}{2}))}{h} + \sin(a+h))$
I dont know where to go from here, how do I simplify to remove the $h$ in the denominator of the left fraction
\begin{align*} \frac{F(x+h)-F(x)}{h} & = \frac{(x+h)[\sin x\cos h +\sin h\cos x]-x\sin x}{h}\\ & =\frac{\cos h -1}{h}x\sin x+\frac{\sin h}{h}x\cos x+\sin x\cos h+\sin h \cos x \end{align*} Now apply $\lim_{h\to 0}$ to both sides, remembering that $\lim_{h\to 0} \frac{\sin h}{h}=1$, and you should get your answer.
If you want to calculate $\lim_{h\to 0}\frac{\cos h-1}{h}$ just multiply the numerator and denominator by the conjugate, and remember the previous identity. Spoiler alert: The answer is $0$.