Question

how to find tangent line of y=cosx-sinx at (pi, -1)

Answer 1

since the point is on the curve, find dy/dx using (pi,-1) to find the equation of the line

Answer 2

Take the derivative y'=-sinx-cosx, evaluate at (pi,-1)

y' at (pi,-1)= 0 - (-1) = 1 = slope m of the tangent line

Point-slope

y-(-1)=(1)(x-pi)

Therefore

y=x-pi-1

Answer 3

lim {[(cosx+h)-sin(x+h)]-[cosx-sinx]}/h as h->0