Question

Найти производную сложной функции y=ln*sqrt(1+tg^2x)

Answer 1

$(ln \sqrt{1+tg^2x})'= \frac{1}{\sqrt{1+tg^2x}} *(\sqrt{1+tg^2x} )'= \\ \\ = \frac{1}{\sqrt{1+tg^2x}} *\frac{1}{2\sqrt{1+tg^2x}} *(1+tg^2x )'= \\ \\= \frac{1}{\sqrt{1+tg^2x}} *\frac{1}{2\sqrt{1+tg^2x}} *2tgx*(tgx )'= \\ \\ = \frac{1}{\sqrt{1+tg^2x}} *\frac{1}{2\sqrt{1+tg^2x}} *2tgx* \frac{1}{cos^2x} = \\ \\ =\frac{2tgx}{2\sqrt{1+tg^2x}*\sqrt{1+tg^2x}*cos^2x} =\frac{tgx}{cos^2x(1+tg^2x)} =\frac{tgx}{cos^2x+sin^2x} =tgx$

Answer 2

$y=ln\sqrt{1+tg^2x}\\\\y'=\frac{1}{\sqrt{1+tg^2x}}\cdot \frac{1}{2\sqrt{1+tg^2x}}\cdot 2tgx\cdot \frac{1}{cos^2x}= \frac{tgx}{cos^2x(1+tg^2x)}=\\\\= \frac{tgx}{cso^2x\cdot \frac{1}{cos^2x}}=tgx$