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Mathematics 8 Online
OpenStudy (anonymous):

derivation of xsinx/(1+tanx)?please!

OpenStudy (amistre64):

deriving is easy, its integrating that that you dont want to meet with in a dark classroom

OpenStudy (amistre64):

use quotient rule \[\frac{(1+tan(x))\ (x\ sin(x))'-(1+tan(x))'\ (x\ sin(x))}{(1+tan(x))^2}\]

OpenStudy (anonymous):

i need the step!please

OpenStudy (amistre64):

and product rule .... since the top needs a deriving of a product.. \[\frac{(1+tan(x))\ (sin(x)+x\ sin(x)')-(sec^2(x))\ (x\ sin(x))}{(1+tan(x))^2}\] \[\frac{(1+tan(x))\ (sin(x)+x\ cos(x))-(sec^2(x))\ (x\ sin(x))}{(1+tan(x))^2}\]

OpenStudy (amistre64):

those are the steps

OpenStudy (amistre64):

id leave it at that, since trying to go any further would just be a waste of time and effort :)

OpenStudy (amistre64):

to clear the cobwebs, lets strip it down: \[[\frac{ab}{c}]'=\frac{c(ab')-(ab)c'}{c^2}=\frac{c(a'b+ab')-(ab)c'}{c^2}\]

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