Derivative Homework Help: Calculating Limits with F'(x) and f[p(x)]

[matematikuvol]

Homework Statement

Calculate

\lim_{h\to 0}\frac{F[p(x)+hp'(x)]-F[p(x)]}{h}

where F'=f

Homework Equations

\lim_{h\to 0}\frac{F(x+h)-F(x)}{h}=F'(x)

The Attempt at a Solution

I think that solution is p'(x)f[p(x)] but I have a trouble to get the result.

 

[namu]

What is $p'(x)$?

It is

\lim_{h \to 0} \frac{p(x+h)-p(x)}{h}

Hence,

p(x)+hp'(x)=p(x+h)

So now we have

\lim_{h\to 0}\frac{F(p(x+h))-F(p(x))}{h}=F'(p(x))=p'(x)f(p(x))

 

[Ray Vickson]

What is $p'(x)$?

It is

\lim_{h \to 0} \frac{p(x+h)-p(x)}{h}

Hence,

p(x)+hp'(x)=p(x+h)

So now we have

\lim_{h\to 0}\frac{F(p(x+h))-F(p(x))}{h}=F'(p(x))=p'(x)f(p(x))

The first result p(x)+hp'(x)=p(x+h) is incorrect, although it can be fixed up. The last line that writes \lim_{h\to 0}\frac{F(p(x+h))-F(p(x))}{h}=F'(p(x)) is also incorrect; it should be dF(p(x))/dx, not F'(p(x)), because, in fact, F'(p(x)) = f(p(x).

A much easier approach is: let R(x,h) = \frac{F[p(x)+hp'(x)]-F[p(x)]}{h}.
(i) if p'(x) = 0 (which can, perhaps, happen at a given point x), then for all h \neq 0, R(x,h) = 0, so the limit is zero, as is p'(x) f(p(x)). (ii) if p'(x) \neq 0,, let k = h p'(x), so that
R(x,h) = p'(x) \frac{F[p(x)+k)-F[p(x)]}{k}. As h \rightarrow 0 we have also that k \rightarrow 0, so the limit is p'(x) f(p(x)).

RGV