anonymous
  • anonymous
Differentiate the function y = ln |2-x-5x^2|
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
is the ques correct?
anonymous
  • anonymous
yup
anonymous
  • anonymous
oops

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anonymous
  • anonymous
fixed it
anonymous
  • anonymous
...
amistre64
  • amistre64
\[[ln(u)]'=\frac{u'}{u}\]
anonymous
  • anonymous
what are those brackets or ||
anonymous
  • anonymous
because it's not |lnu| it's ln|u|
amistre64
  • amistre64
those are just for grouping; it means that the whole of it is derived and not just the (u) part
anonymous
  • anonymous
okay so show me how
amistre64
  • amistre64
i believe it goes piece wise in the end
amistre64
  • amistre64
or you can use the sqrt(u)^2 trick
anonymous
  • anonymous
please show me
anonymous
  • anonymous
|dw:1334421083017:dw|
amistre64
  • amistre64
\[\frac{d}{dx}ln(\sqrt{u\ }^2)=\frac{(\sqrt{u\ }^2)'}{ \sqrt{u\ }^2}\to\ \frac{2\sqrt{u\ }*\sqrt{u}'}{ \sqrt{u\ }^2}\]
amistre64
  • amistre64
if i follow that derivatives i think it would look like this ... but best to check with the wolf
amistre64
  • amistre64
http://www.wolframalpha.com/input/?i=derivative+ln+%7C2-x-5x%5E2%7C the wolf doesnt seem to go thru the hoopla tho
amistre64
  • amistre64
\[\frac{2\sqrt{u\ }}{ \sqrt{u\ }^2}\sqrt{u}'\] \[\frac{2\sqrt{u\ }}{ \sqrt{u\ }^2}\frac{u'}{2\sqrt{u}}=\frac{u'}{ \sqrt{u\ }^2}\to\ \frac{u'}{ |u|}\]
anonymous
  • anonymous
whatevr it's 1/(2-x-5x^2) * -(10x+1) = 10x+1 / 5x^2+x-2 woohoo

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