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
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"#