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)]#