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

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Calc2.62

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