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jverzani
2024-07-31 11:24:53 -04:00
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quarto/derivatives/derivatives.qmd
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@@ -730,13 +730,13 @@ $$
\frac{f(g(x+h)) - f(g(x))}{h} = \frac{f(g(x+h)) - f(g(x))}{g(x+h) - g(x)} \cdot \frac{g(x+h) - g(x)}{h}.
$$
The left hand side will converge to the derivative of $u(x)$ or $[f(g(x))]'$.
The left-hand side will converge to the derivative of $u(x)$ or $[f(g(x))]'$.
The right most part of the right side would have a limit $g'(x)$, were we to let $h$ go to $0$.
The right-most part of the right-hand side would have a limit $g'(x)$, were we to let $h$ go to $0$.
It isn't obvious, but the left part of the right side has the limit $f'(g(x))$. This would be clear if *only* $g(x+h) = g(x) + h$, for then the expression would be exactly the limit expression with $c=g(x)$. But, alas, except to some hopeful students and some special cases, it is definitely not the case in general that $g(x+h) = g(x) + h$ - that right parentheses actually means something. However, it is *nearly* the case that $g(x+h) = g(x) + kh$ for some $k$ and this can be used to formulate a proof (one of the two detailed [here](http://en.wikipedia.org/wiki/Chain_rule#Proofs) and [here](http://kruel.co/math/chainrule.pdf)).
It isn't obvious, but the left part of the right-hand side has the limit $f'(g(x))$. This would be clear if *only* $g(x+h) = g(x) + h$, for then the expression would be exactly the limit expression with $c=g(x)$. But, alas, except to some hopeful students and some special cases, it is definitely not the case in general that $g(x+h) = g(x) + h$ - that right parentheses actually means something. However, it is *nearly* the case that $g(x+h) = g(x) + kh$ for some $k$ and this can be used to formulate a proof (one of the two detailed [here](http://en.wikipedia.org/wiki/Chain_rule#Proofs) and [here](http://kruel.co/math/chainrule.pdf)).
Combined, we would end up with: