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 Exercise [06.07] 
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Joined: 11 May 2008, 20:07
Posts: 11
Location: france
Post Exercise [06.07]
Let's calculate d(\frac{f(x)}{g(x)})
The tools we've got are the relations :

d(f(x)g(x))=g(x)df(x)+f(x)dg(x)<br /><br />d{f(g(x))} = f'(g(x))g'(x)<br /><br />d(x^{n}) = nx^{n-1}dx

Now we can make it :

d(\frac{f(x)}{g(x)})=d(f(x){g(x)}^{-1})={g(x)}^{-1}d(f(x))+f(x)d({g(x)}^{-1})

Let's introduce another function :

h(x)=x^{-1}<br /><br />{g(x)}^{-1}=h(g(x))<br /><br />d(h(x))=-x^{-2}<br /><br />d({g(x)}^{-1})=d{h(g(x))}=-dg(x){g(x)}^{-2}

Now we can find :

d(\frac{f(x)}{g(x)})={g(x)}^{-1}d(f(x))-f(x)dg(x){g(x)}^{-2}= \frac{g(x)df(x)-f(x)dg(x)}{{g(x)}^2}


13 May 2008, 08:09
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Joined: 07 Jun 2008, 08:21
Posts: 235
Post Re: Exercise [06.07]
I think your proof contains a few mistakes. Here is my version.
It is important to understand that:

f'(g(x))means \frac{df}{dg} and that f'(x) means \frac{df}{dx}

We are asked to use the relations :

d(f(x)g(x))=g(x)df(x)+f(x)dg(x) --------1

d{f(g(x))} = f'(g(x))g'(x) -------------------2

d(x^{n}) = nx^{n-1}dx --------------3

Using the Leibnitz rule (1) gives:

d(\frac{f(x)}{g(x)})=d(f(x){g(x)}^{-1})={g(x)}^{-1}d(f(x))+f(x)d({g(x)}^{-1}) --------------4

At this point we need to find an expression for d({g(x)}^{-1})

Let h(g(x))=g(x)^{-1}

Using the relation 2 above

d\{h(g(x))\}=h'(g(x))g'(x) -----------------5

using relation 3 above

h'(g(x))=-g(x)^{-2}

So in 4

d(g(x)^{-1})=d\{h(g(x))\}=h'(g(x))g'(x)=-g(x)^{-2}g'(x)=-g(x)^{-2}dg(x)

Substituting into 4 gives

d(\frac{f(x)}{g(x)})={g(x)}^{-1}d(f(x))-f(x)dg(x){g(x)}^{-2}= \frac{g(x)df(x)-f(x)dg(x)}{{g(x)}^2}


07 Jul 2008, 19:45
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