Question

Question #b3476

Answer 1

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

Explanation:

`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)]`