Question

find dy/dx if y= (tanx)/ (1+sinx)

Answer 1

\[vu'-uv'/v^2\]

Answer 2

dy/dx = [(1+sinx)(sec^2) - (tanx)(cosx)] / (1+sinx)^2

Expand and simplify.

Answer 3

\[tan(x)(1+sinx)^{-1}=tan(x)-1(1+sinx)^{-2}(cosx)+(1+sinx)^{-1}(sec^2x)\]

Answer 4

\[\frac{-tanxcosx}{(1+sinx)^2}+\frac{sec^2x}{1+sinx}\]

Answer 5

\[\frac{-tanxcosx}{(1+sinx)^2}+\frac{sec^2x(1+sinx)}{(1+sinx)^2}\]

Answer 6

\[\frac{-tanxcosx+sec^2x+sec^2xsinx}{(1+sinx)^2}\]

Answer 7

\[-tanxcosx=\frac{-sinx}{cosx}(cosx)=-sinx\]

Answer 8

you can do more if you want like switching them to have all sin(x) in them using the trig identity

\[cos^2x+sin^2x=1\]

Answer 9

thanks, i appreciate it =]