Question #b3476

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Answer 1 · 2017-06-12T02:59:29

$g'(x)=-tanxcosx["lin"(cosx)]$

$g(x)=sin["lin"(cosx)]$

Let $u="lin"(cosx)$

And $v=cosx$

Then $g(x)=sinu$

And $u="lin"v$

And $g'(x)=d/dx(sinu)xx(du)/dx xx (dv)/dx$

$d/dx(sinu)=cosu=cos("lin"v)=cos["lin"(cosx)]$

$(du)/dx=1/v=1/cosx$

$(d v)/dx=-sinx$

$g'(x)=cos["lin"(cosx)]xx1/cosx xx -sinx$

$=-tanxcosx["lin"(cosx)]$