How do I find the equation to the tangent linf of f(x)=1+sinx/1-sinx at (pi,1)?

1 Answer
Feb 20, 2018

#y=-2x+(2pi+1)#

Explanation:

#•color(white)(x)m_(color(red)"tangent")=f'(pi)#

#"differentiate using the "color(blue)"quotient rule"#

#"given "f(x)=(g(x))/(h(x))" then"#

#f'(x)=(h(x)g'(x)-g(x)h'(x))/(h(x))^2larrcolor(blue)"quotient rule"#

#g(x)=1+sinxrArrg'(x)=cosx#

#h(x)=1-sinxrArrh'(x)=-cosx#

#rArrf'(x)=(cosx(1-sinx)+cosx(1+sinx))/(1-sinx)^2#

#color(white)(rArrf'(x))=(cosx(1-sinx+1+sinx))/(1-sinx)^2#

#color(white)(rArrf'(x))=(2cosx)/(1-sinx)^2#

#rArrf'(pi)=-2/(1)^2=-2#

#rArry-1=-2(x-pi)#

#rArry-1=-2x+2pi#

#rArry=-2x+(2pi+1)larrcolor(red)"equation of tangent"#