How do you find the equation of the line tangent to #y=cosx-sinx#, at (pi,-1)?
1 Answer
Jul 30, 2017
Explanation:
#•color(white)(2/2)dy/dx=m_(color(red)"tangent")" at x = a"#
#y=cosx-sinx#
#rArrdy/dx=-sinx-cosx#
#x=pitody/dx=-sinpi-cospi=1#
#rArry+1=x-pi#
#rArry=x-(pi+1)larr" equation of tangent"#
