Deriving the derivative

Question

I have the following expression:

$$ \frac{W\left(\frac{p}{1-\gamma\Delta}\right) - W(p)}{\Delta}$$

and want to find out what happens as $\Delta \to 0$. I reformulate as

$$ W\left(\frac{p}{1-\gamma\Delta}\right) - W(p) \\ = W\left(p - p + \frac{p}{1-\gamma\Delta}\right) - W(p) \\ = W\left(\frac{-p(1-\gamma\Delta)}{1 - \gamma\Delta} + \frac{p}{1-\gamma\Delta} + p\right) - W(p) \\ = W\left(\frac{\gamma\Delta p}{1 - \gamma\Delta} + p\right) - W(p) $$

and then

$$\lim_{\Delta\to 0}\frac{W\left(\Delta \frac{\gamma p}{1-\gamma\Delta}+p\right)-W(p)}{\Delta}=W'(p)\gamma p$$

Is this correct? The last step is mostly intuition, and I'd be happy if someone could provide some more foundation here.

Answer 1

Yes. Another way to get your answer: Let $f(\Delta) := \frac p{1-\gamma \Delta} $. Then $f(0)=p,$ and so your fraction is $$ \frac { W(\frac p{1-\gamma \Delta}) - W(p) }{\Delta} = \frac{W(f(\Delta)) - W(f(0))}{\Delta}, $$ and we recognize a difference quotient for $W\circ f$. By chain rule,

$$ \lim_{\Delta \to 0} \frac{W(f(\Delta)) - W(f(0))}{\Delta} = (W\circ f)'(0) = W'(f(0)) f'(0) =W'(p)\gamma p.$$

Answer 2

$$\lim_{\Delta\rightarrow 0}\frac{W\left(\frac{p}{1-\gamma\Delta}\right)-W(p)}{\Delta}$$

Your calculation is correct.

But it is not necessary to transform the numerator of your expression. It is obvious that

$$\lim_{\Delta\rightarrow 0}\left(W\left(\frac{p}{1-\gamma\Delta}\right)-W(p)\right)=0.$$

So you have an expression $\frac{0}{0}.$

Apply simply L'Hôpital's rule for $\frac{0}{0}$: If $\lim_{x\rightarrow 0}f'(x)$ and $\lim_{x\rightarrow 0}g'(x)$ exist, then $\lim_{x\rightarrow 0}\frac{f(x)}{g(x)}=\frac{\lim_{x\rightarrow 0}f'(x)}{\lim_{x\rightarrow 0}g'(x)}$.

$$\lim_{\Delta\rightarrow 0}\frac{W\left(\frac{p}{1-\gamma\Delta}\right)-W(p)}{\Delta}=\frac{\lim_{\Delta\rightarrow 0}W'(\frac{p}{1-\gamma\Delta})p\gamma}{\lim_{\Delta\rightarrow 0}(1+\gamma\Delta)^2}$$

$$=W'(p)\gamma p$$