Answer to Question #167834 in Calculus for Vishal

Question

Find f'(x) of f(x)= 1/{x-[2/(x + sinx)]}

Answer 1

$f(x)=1/(x-2/(x+sinx))$

$f`(x)=-1/(x-2/(x+sinx))^2*(x-2/(x+sinx))`$

$(x-2/(x+sinx))`=1+2/(x+sinx)^2(x+sinx)`$

$(x+sinx)`=1+cosx$

$f`(x)=-1/(x-2/(x+sinx))^2*(1+2/(x+sinx)^2*(1+cosx))= ((x+sinx)^2 + 2 +2cosx)/(x(x+sinx)-2)^2$