Calc2.62

From Example Problems
Revision as of 00:20, 27 November 2007 by Scottie 000 (Talk | contribs)

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
Jump to: navigation, search

\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