Question

1) f(x)=x2+2x/x-1
2) f(x)=2cosx-tgx

Answer 1

f'(x) =[ (2x+2)*(x-1) -(x^2+2x)*1 ] / (x-1) ^2 = [ 2x^2-2-x^2-2x ] /(x-1)^2 =

=[ x^2-2x-2 ] / (x-1)^2

f'(x)= 2*(-sinx) - 1/cos^2x = -2sinx - 1/cos^2x