Question 1159793
{{{y=e^x/(1+2e^x )}}}

if {{{x=0}}}

find tangent point:

{{{y=e^0/(1+2e^0 )}}}

{{{y=1/(1+2*1 )}}}

{{{y=1/3}}}

tangent point is: ({{{0}}},{{{1/3}}})



we need a slope of the tangent line, so find derivative:


{{{d/dx)=e^x/(1+2e^x )}}}

apply the Quotient Rule 

{{{y}}}'{{{(x) =((d/dx)e^x*(1+2e^x )- e^x*(d/dx)(1+2e^x ))/(1+2e^x )^2}}}..........this simplifies to

{{{y}}}'{{{(x) = 1/(2 (2e^x + 1)) - 1/(2 (2e^x + 1)^2)}}}


use given {{{x=0 }}}to find the slope of the tangent line


{{{y}}}'{{{(0) = 1/(2 (2e^0 + 1)) - 1/(2 (2e^0 + 1)^2)}}}

{{{y}}}'{{{(0) = 1/(2 (2*1+ 1)) - 1/(2 (2*1 + 1)^2)}}}

{{{y}}}'{{{(0) = 1/6 - 1/18}}}

{{{y}}}'{{{(0) = 1/9}}}=> the slope of the tangent line is {{{m=1/9}}}


tangent line is:

{{{y=mx+b}}}..............use tangent point  ({{{0}}},{{{1/3}}}) and the slope {{{m=1/9}}}

{{{1/3=(1/9)*0+b}}}

{{{b=1/3}}}


and  tangent line is:

 {{{ y=(1/9)x+1/3}}}