Question

f(x)=3x-tgx чему равно f'(0)=?

Answer 1

$F(x)=3x-tgx \\ \\ F'(x)=(3x-tgx)' = x - ( \frac{sinx}{cosx} )' = x - \frac{(sinx)' cosx - sinx(cosx)' }{cos^2x} = \\ \\ =x - \frac{cosx* cosx - sinx(-sinx) }{cos^2x} =x - \frac{cosx^2 + sinx^2}{cos^2x} =x - \frac{1}{cos^2x} \\ \\ F'(0) = 0 - \frac{1}{cos^2 0} =-1$