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Differentation - HSN forum

# Differentation

4 replies to this topic

### #1AM4R

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Posted 19 December 2006 - 09:00 PM

Can someone help with this please. I dont know where to start and what rules I use to get through this question.

ln(1 + x/1 - x)

Thanks
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Posted 19 December 2006 - 09:36 PM

Let , so you have . You can differentiate y with repsect to x using the chain rule now. Note that you will probably use the quotient rule to differentiate u with respect to x.
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### #3dfx

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Posted 20 December 2006 - 12:10 PM

You could algebraically simplify the:

(exclude the u, this is just cause I'm too lazy to rewrite the expression)

to 1/(1-x) which makes it a pretty straightforward integral, ln[1/(1-x)]

edit: sorry I'm pretty sure I read it was an integral question! Anyway you still could simplify and use the chain rule.

### #4AM4R

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Posted 20 December 2006 - 08:19 PM

whoops i didnt make that clear. Its differentiation and Its ln (1+X)/(1-X)

thanks
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### #5waitingforan_alibi

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Posted 20 December 2006 - 11:39 PM

just do:

ln((1+x)/(1-x)= ln(1+x)-ln(1-x)

that differentiates to:

1/(1+x)+1/(1-x)

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