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Question: First Principals - HSN forum

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Question: First Principals


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#1 Floorball Maniac

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Posted 01 October 2005 - 12:46 PM

Derive the following from first principals.

F(x) = e powerx.gif


Can someone please show me how to do it.

Thanks!

#2 Dave

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Posted 01 October 2005 - 12:57 PM

deriving from 1st principles

there is an equation which should be in your notes you just sub in values

i cant actually remember the equation and i binned my notes so not much help other than that

If i am not here i am somewhere else



#3 Steve

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Posted 01 October 2005 - 05:39 PM

The definition of the derivative of f(x) is



\begin{align*}f'(x) = \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\end{align*}
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#4 Floorball Maniac

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Posted 01 October 2005 - 06:33 PM

I have got that part but how do you do the simplifying in order to work it out?

#5 sparky

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Posted 01 October 2005 - 08:10 PM

Okay

x -> e^x

x+h -> e^x+h

f(x+h) - f(x) / h = e^x+h - e^x / h
= e^x . e^h - e^x / h
= e^x (e^h -1) / h
Taking out a common factor

Substitute into formula, Steve stated above
f'(x) = lim h->0 f(x+h) - f(x) / h
= lim h->0 e^x (e^h -1) / h
= e^x . 1
= e^x
Mark

#6 Floorball Maniac

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Posted 02 October 2005 - 10:33 AM

Thanks! I got most of that apart from a couple of lines.

QUOTE(sparky @ Oct 1 2005, 09:10 PM)
      = lim h->0  e^x (e^h -1) / h 
      = e^x . 1

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How did you get that above? I thought you put 0 in place of h but would that not give you 0 since you are dividing by 0?

#7 dfx

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Posted 02 October 2005 - 11:05 AM

Yeah, so as it tends to infinity....

#8 Steve

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Posted 04 October 2005 - 06:57 PM

You can't go straight from f'(x)= \lim_{h \rightarrow 0}  \frac{e^x (e^h -1)}{h} to the answer since taking h to zero would require dividing by zero.

We have to get rid of the h on the bottom line somehow...

Suppose we have:


\begin{align*}
y &= \frac{e^x (e^h -1)}{h} \\
\ln y &= \ln{\left( \frac{e^x (e^h -1)}{h} \right)} \\
\ln y &= \ln{\left( e^x (e^h -1)}\right) - \ln{h} \qquad \textrm{ by the laws of logs}\\
\ln y &= \ln{e^x} + \ln{(e^h -1)} - \ln{h} \\
\ln y &= x + \ln{0} - \ln{0} \qquad \textrm{ if } h=0 \\
\end{align*}

So y=f'(x)=e^x.

Hope that helps in some way smile.gif
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#9 Floorball Maniac

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Posted 05 October 2005 - 05:13 PM

Great help, thanks!





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