- #1

"Don't panic!"

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From the definition of the derivative of a differentiable function [itex]f:\mathbb{R}\rightarrow\mathbb{R}[/itex] (one-dimensional case), we have that [tex]f'(x)=\frac{df}{dx}=\lim_{\Delta x\rightarrow 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}= \lim_{\Delta x\rightarrow 0}\frac{\Delta f}{\Delta x}[/tex]

This implies that [tex]\frac{\Delta f}{\Delta x}= f'(x) +\varepsilon (x)\quad\Rightarrow\quad\Delta f = \left(f'(x)+\varepsilon (x)\right)\Delta x [/tex]

where [itex]\varepsilon (x)[/itex] is some error function which accounts for the difference between the actual (finite) change in [itex]f[/itex] and its linear approximation [itex]f'(x)\Delta x[/itex]. Furthermore, [itex]\varepsilon (x)[/itex] satisfies the property [tex]\lim_{\Delta x\rightarrow 0}\varepsilon (x)=0[/tex] such that as [itex]\Delta x \rightarrow 0,\quad\frac{\Delta f}{\Delta x}\rightarrow f'(x)[/itex].

Now, consider a function [itex]y=f\circ g(x)=f(g(x))[/itex] and let [itex]u=g(x)[/itex]. We have then, that [tex]\Delta u = g(x+\Delta x)-g(x)=\left(g'(x)+\varepsilon_{1}(x)\right)\Delta x[/tex] [tex]\Delta y = f(u+\Delta u)-f(u)=\left(f'(u)+\varepsilon_{2}(u)\right)\Delta u[/tex]

Note that [itex]\varepsilon_{2}(u)\rightarrow 0[/itex] as [itex]\Delta u\rightarrow 0[/itex]. However, since [itex]\Delta u\rightarrow 0[/itex] as [itex]\Delta x\rightarrow 0[/itex], this implies that [itex]\varepsilon_{2}(u)\rightarrow 0[/itex] as [itex]\Delta x\rightarrow 0[/itex].

And so,

[tex]f(u+\Delta u)-f(u)=\left(f'(u)+\varepsilon_{2}(u)\right)\Delta u[/tex] [tex]\Rightarrow f(g(x+\Delta x))-f(g(x))=f\circ g(x+\Delta x)-f\circ g(x)\\ \qquad\qquad\qquad\qquad\qquad\quad\;\;=\left(f'(g(x))+\varepsilon_{2}(g(x))\right)\cdot\left(g'(x)+\varepsilon_{1}(x)\right)\Delta x\\ \qquad\qquad\qquad\qquad\qquad\quad\;\;=f'(g(x))f'(g(x))g'(x)\Delta x +\left(f'(g(x)) \varepsilon_{1}+g'(x)\varepsilon_{2}+\varepsilon_{1}\varepsilon_{2}\right)\Delta x\\ \qquad\qquad\qquad\qquad\qquad\quad\;\;=f'(g(x))g'(x)\Delta x +\varepsilon_{3}\Delta x[/tex]

where [itex]\varepsilon_{3}\equiv f'(g(x)) \varepsilon_{1}+g'(x)\varepsilon_{2}+\varepsilon_{1}\varepsilon_{2}[/itex]. We see from this that as [itex]\Delta x\rightarrow 0,\quad\varepsilon_{3}\rightarrow 0[/itex]. Hence,

[tex]\lim_{\Delta x\rightarrow 0}\frac{f\circ g(x+\Delta x)-f\circ g(x)}{\Delta x}= (f\circ g)'(x)=f'(g(x))g'(x)[/tex]