Question

@Michele_Laino

Answer 1

please you have to apply the chain rue, i order to that, i write your function as below:
\[y=\arctan z, z=2x\]
now the derivative of y=arctan(z), is:
\[y'=\frac{ 1 }{ 1+z ^{2} }\]
so? please complete the calculus

Answer 2

so it is 1/1+4^2

Answer 3

wait no it is 2/1+2x^2

Answer 4

Sorry I think is:
\[f'(x)=\frac{ 2 }{ 1+4x ^{2} }\]

Answer 5

ohh

Answer 6

thats what i meant