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Differentiation/integration


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#21 ad absurdum

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Posted 21 April 2006 - 02:25 PM

QUOTE(dfx @ Apr 21 2006, 03:02 PM) View Post

QUOTE(George @ Apr 20 2006, 11:00 PM) View Post

(have you ever tried it with sinx? smile.gif )


Oh yes I was recently challenged by my maths teacher to differentiate from first principles. I believe it involves the taylor and maclaurin expansions for Sinx... but yeah I gave up lol.
You can do it with higher maths material. The only identities you need are:


\begin{align*}\sin(x+h) = \sin(x)\cos(h) + \cos(x)\sin(h)\end{align*}


\begin{align*}\sin^2(x) = 1 - \cos^2(x)\end{align*}

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#22 dfx

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Posted 21 April 2006 - 03:03 PM

Yes the but I'm pretty sure you still need the taylor and maclaurin series... however I stand corrected! smile.gif

#23 ad absurdum

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Posted 21 April 2006 - 04:54 PM

I'm not even going to pretend I know what they are tongue.gif
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#24 George

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Posted 22 April 2006 - 01:21 PM

Here's the way I approached it:



\begin{align*}
 \frac{d}{{dx}}\sin x &= \mathop {\lim }\limits_{h \to 0} \frac{{\sin \left( {x + h} \right) - \sin x}}{h} \\ 
  &= \mathop {\lim }\limits_{h \to 0} \frac{{\sin x\cos h  + \cos x\sin h  - \sin x}}{h} \\ 
  &= \mathop {\lim }\limits_{h \to 0} \left( {\sin x\frac{{\cos h  - 1}}{h} + \cos x\frac{{\sin h }}{h}} \right) \\ 
  &= \sin x.0 + \cos x.1 \\ 
  &= \cos x 
\end{align*}

However, that relies on two facts, 

\begin{align*}
  \mathop {\lim }\limits_{h \to 0} \frac{{\cos h  - 1}}{h} = 0
\end{align*}

and 

\begin{align*}
  \mathop {\lim }\limits_{h \to 0} \frac{\sin h}{h} = 1
\end{align*}

. The sinx / x limit requires a lot of work if you're being rigorous, and the cos limit follows from that quite easily.

Now, in practice, nobody would do all that work every time they needed to differentiate sinx!

The whole point of maths is to come up with theorems and rules. Once they're proved, it makes sense to just use the result - there's no point reinventing the wheel all the time!

#25 dfx

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Posted 22 April 2006 - 02:43 PM

I was puzzling over why you delved into Cosh and Sinh when I realized it was h for height. laugh.gif

#26 ad absurdum

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Posted 23 April 2006 - 12:33 PM

QUOTE(George @ Apr 22 2006, 02:21 PM) View Post

Here's the way I approached it:



\begin{align*}
 \frac{d}{{dx}}\sin x &= \mathop {\lim }\limits_{h \to 0} \frac{{\sin \left( {x + h} \right) - \sin x}}{h} \\ 
  &= \mathop {\lim }\limits_{h \to 0} \frac{{\sin x\cos h  + \cos x\sin h  - \sin x}}{h} \\ 
  &= \mathop {\lim }\limits_{h \to 0} \left( {\sin x\frac{{\cos h  - 1}}{h} + \cos x\frac{{\sin h }}{h}} \right) \\ 
  &= \sin x.0 + \cos x.1 \\ 
  &= \cos x 
\end{align*}

However, that relies on two facts, 

\begin{align*}
  \mathop {\lim }\limits_{h \to 0} \frac{{\cos h  - 1}}{h} = 0
\end{align*}

and 

\begin{align*}
  \mathop {\lim }\limits_{h \to 0} \frac{\sin h}{h} = 1
\end{align*}

. The sinx / x limit requires a lot of work if you're being rigorous, and the cos limit follows from that quite easily.

Now, in practice, nobody would do all that work every time they needed to differentiate sinx!

The whole point of maths is to come up with theorems and rules. Once they're proved, it makes sense to just use the result - there's no point reinventing the wheel all the time!
This is also how I done it, except I didn't know how to calculate the limits myself and had to cheat a little with them.

dfx, h is meant to represent a small change in x, that's why the differentiated function is f(x+h) - f(x) / h - i.e. the change in f(x) over a change in x. Obviously the most accurate formula for the gradient will come as h is closer to zero.
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#27 dfx

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Posted 23 April 2006 - 04:15 PM

hehe yep.

#28 Vyka

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Posted 28 April 2006 - 01:45 PM

QUOTE(Daniel Williamson @ Mar 27 2006, 08:51 PM) View Post

Any easy ways to remember the difference between the two? add one to power divide by new power i mean come on ph34r.gif




PRACTICE!!! Just like the rest of us hahaha

#29 ScotlandGirl

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Posted 30 April 2006 - 07:33 PM

This thread scares me blink.gif

And actually there is a semi-easy way to remember the difference (well I think so anyway!). For integration you increase the power and for differentiation you decrease the power.





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