namu said:
What is ##p'(x)##?
It is
[tex]\lim_{h \to 0} \frac{p(x+h)-p(x)}{h}[/tex]
Hence,
[tex]p(x)+hp'(x)=p(x+h)[/tex]
So now we have
[tex]\lim_{h\to 0}\frac{F(p(x+h))-F(p(x))}{h}=F'(p(x))=p'(x)f(p(x))[/tex]
The first result [tex]p(x)+hp'(x)=p(x+h)[/tex] is incorrect, although it can be fixed up. The last line that writes [tex]\lim_{h\to 0}\frac{F(p(x+h))-F(p(x))}{h}=F'(p(x))[/tex] is also incorrect; it should be [itex]dF(p(x))/dx[/itex], not [itex]F'(p(x))[/itex], because, in fact, [itex]F'(p(x)) = f(p(x).[/itex]
A much easier approach is: let [tex]R(x,h) = \frac{F[p(x)+hp'(x)]-F[p(x)]}{h}.[/tex]
(i) if p'(x) = 0 (which can, perhaps, happen at a given point x), then for all [itex]h \neq 0,[/itex] R(x,h) = 0, so the limit is zero, as is [itex]p'(x) f(p(x)).[/itex] (ii) if [itex]p'(x) \neq 0,[/itex], let [itex]k = h p'(x)[/itex], so that
[tex]R(x,h) = p'(x) \frac{F[p(x)+k)-F[p(x)]}{k}.[/tex] As [itex]h \rightarrow 0[/itex] we have also that [itex]k \rightarrow 0[/itex], so the limit is [itex]p'(x) f(p(x)).[/itex]
RGV