Question 1156570


{{{y = (tan^-1(4x))^2}}}

apply the chain rule : {{{df(u)/dx=(df/du)* (du/dx)}}}

where {{{f=u^2}}} and  {{{u=tan^-1(4x)}}}


{{{df(u)/dx=(d/du)(u^2)* (du/dx)(tan^-1(4x))}}}

{{{(d/du)(u^2)=2u}}}

{{{(du/dx)(tan^-1(4x))=4/(16x^2+1)}}}


{{{df(u)/dx=y}}}'


{{{y}}}'={{{2u* (4/(16x^2+1))}}}


substitute back {{{u=tan^-1(4x)}}}


{{{y}}}'={{{2tan^-1(4x)* (4/(16x^2+1))}}}


{{{y}}}'={{{(8tan^-1(4x))/(16x^2+1))}}}