Question

Найти дифференциал функции:y=sqrt(1+x^2)arctg x

Answer 1

dy=f'(x)dx

$f'(x)=(\sqrt{1+x^2}*arctgx)'=\\=(\sqrt{1+x^2})'*arctgx+\sqrt{1+x^2}*(arctgx)'=\\=\frac{1}{2\sqrt{1+x^2}}*(1+x^2)'*arctgx+\sqrt{1+x^2}*\frac{1}{1+x^2}}=\\=\frac{x*arctgx}{\sqrt{1+x^2}}+\frac{1}{\sqrt{1+x^2}}=\frac{x*arctgx+1}{\sqrt{1+x^2}}\\\\dy=\frac{x*arctgx+1}{\sqrt{1+x^2}}dx$