# Calc2.62

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$\left(f(g(x))\right)'=f'(g(x))g'(x)$

$\Rightarrow \left|{\frac {f(g(x+\delta _{{fg}}))-f(g(x))}{\delta _{{fg}}}}-f'(g(x))g'(x)\right|<\varepsilon _{{fg}}\,$, where subscripts denote dependence. (Note all $\delta \,$ and $\varepsilon \,$ are positive )

We need to show the above follows from the differentiability of f and g:

$\left|{\frac {g(x+\delta _{{g}})-g(x)}{\delta _{{g}}}}-g'(x)\right|<\varepsilon _{{g}}\,$ and $\left|{\frac {f(x+\delta _{{f}})-f(x)}{\delta _{{f}}}}-f'(x)\right|<\varepsilon _{{f}}\,$

On rearranging we get:

$|g(x+\delta _{g})|<\;\varepsilon _{g}\delta _{g}-\delta _{g}\,|g'(x)|-|g(x)|$

$|f(x+\delta _{f})|<\;\varepsilon _{f}\delta _{f}-\delta _{f}\,|f'(x)|-|f(x)|$

Consider $|f(g(x+\delta _{{fg}}))|\,$.

By definition of g, and then f: $|f(g(x+\delta _{{fg}}))|<\left|f\left(\varepsilon _{g}\delta _{{fg}}-\delta _{{fg}}\,|g'(x)|-|g(x)|\right)\right|=\left|f\left(|g(x)|+\delta _{{fg}}\,|g'(x)|-\varepsilon _{g}\delta _{{fg}}\right)\right|$

$<\varepsilon _{f}(\delta _{{fg}}\,|g'(x)|-\varepsilon _{g}\delta _{{fg}})-(\delta _{{fg}}\,|g'(x)|-\varepsilon _{g}\delta _{{fg}})|f'(g(x))|-|f(g(x))|$ , where we are taking $\delta _{f}=\delta _{{fg}}\,|g'(x)|-\varepsilon _{g}\delta _{{fg}}\left(*\right)$

Thus: $|f(g(x+\delta _{{fg}}))|\;+\;|f(g(x))|<\varepsilon _{f}(\delta _{{fg}}\,|g'(x)|-\varepsilon _{g}\delta _{{fg}})-(\delta _{{fg}}\,|g'(x)|-\varepsilon _{g}\delta _{{fg}})|f'(g(x))|$

$\Rightarrow \left|{\frac {f(g(x+\delta _{{fg}}))-f(g(x))}{\delta _{{fg}}}}\right|<\varepsilon _{f}(|g'(x)|-\varepsilon _{g})-(|g'(x)|-\varepsilon _{g})|f'(g(x))|$

$\Rightarrow \left|{\frac {f(g(x+\delta _{{fg}}))-f(g(x))}{\delta _{{fg}}}}\right|+|f'(g(x))||g'(x)|<\varepsilon _{f}(|g'(x)|-\varepsilon _{g})+\varepsilon _{g}|f'(g(x))|$

$\Rightarrow \left|{\frac {f(g(x+\delta _{{fg}}))-f(g(x))}{\delta _{{fg}}}}-f'(g(x))g'(x)\right|<\varepsilon _{f}(|g'(x)|-\varepsilon _{g})+\varepsilon _{g}|f'(g(x))|$

Take $\varepsilon _{{fg}}=\varepsilon _{f}(|g'(x)|-\varepsilon _{g})+\varepsilon _{g}|f'(g(x))|$

From $(*)\,$ we see that as $\delta _{{fg}}\to 0,\delta _{f}\to 0\Rightarrow \varepsilon _{f}\to 0$
We also have $\delta _{g}\to 0\Rightarrow \varepsilon _{g}\to 0$
Hence $\varepsilon _{{fg}}\to 0$

Therefore $\left|{\frac {f(g(x+\delta _{{fg}}))-f(g(x))}{\delta _{{fg}}}}-f'(g(x))g'(x)\right|<\varepsilon _{{fg}}\,$ with $\varepsilon _{{fg}}\to 0$ as $\delta _{g}\to 0$, which is what we set out to prove