Question

Вычислите производную y=ln arctgкорень1+xв квадрате

Answer 1

$(\ln{arctg(\sqrt{1+x^2})'=\frac{(arctg(\sqrt{1+x^2})'}{arctg(\sqrt{1+x^2})}=\frac{\frac{1}{1+x^2 +1}\cdot(\sqrt{1+x^2})'}{arctg(\sqrt{1+x^2})}$$\\ \\ = \frac{2x \cdot \frac{1}{2 \cdot \sqrt{1+x^2}}}{(x^2 +2)\cdot arctg(\sqrt{1+x^2})} = \frac{x}{\sqrt{1+x^2} \cdot (x^2 +2)\cdot arctg(\sqrt{1+x^2})}$

Answer 2

$y`=1/arctg \sqrt{1+x^2} *1/(1+(1+x^2))*2x/2 \sqrt{1+x^2} =x/(2+x^2)$$\sqrt{1+x^2} arctg \sqrt{x+x^2}$