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