Помогите решить пожалуйста

F(x) = 4x³ + 6x + 3
f '(x) = (4x³)' + (6x)' + 3' = 12x² + 6
f '(1) = 12 * 1² + 6 = 12 + 6 = 18

$f(x) = \frac{x}{1+ x^{2} } \\\\f'(x) = \frac{x' *(1+ x^{2} )-x*(1+ x^{2} )'}{(1+ x^{2} ) ^{2} } = \frac{1+ x^{2} -x*2x}{(1+ x^{2} ) ^{2} } = \frac{1+ x^{2} -2 x^{2} }{(1+ x^{2} ) ^{2} }= \frac{1- x^{2} }{(1+ x^{2} ) ^{2} }\\\\f'(0) = \frac{1-0 ^{2} }{(1+0 ^{2}) ^{2} } = \frac{1}{1} =1$