## anonymous one year ago IS ANYOME GOOD AT FINDING DERIATIVES FOR LN?! Please I need help as soon as possible! :(

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1. alekos

post it up

2. anonymous

Ok

3. anonymous

y=ln(1-x/(x+2)^5)

4. alekos

$y=\ln [1 -\frac{ x }{(x+2)^{5}}]$ Is that it?

5. anonymous

No sorry my computer is messed up, the 1-x is the numerator, and the denominator is as you out it. Ln is correct too

6. anonymous

Put*

7. anonymous

@alekos

8. IrishBoy123

$$\large y=ln \frac{1-x}{(x+2)^5}$$ assuming that you know that $$\large (log \ u(x))' = \frac{1}{u} u'$$ then look also at $$\large ln \frac{a}{b} = ln \ a - ln \ b$$ and similar properties of logs; and maybe split this up a bit before you start...

9. alekos

yeah, thanks Irish Boy. I reckon do the properties of logs.... Ln(1-x) - Ln(x+2)^5 = Ln(1-x) - 5Ln(x+2) the rest is easy because the derivative of Lnx is 1/x

10. IrishBoy123

@alekos look good