Найти производную функции \(f(x) = x+(x-1)\arcsin\sqrt{\frac{x}{x+1}}\).
Решение.
\(\begin{multline}
f'(x) = \left(x+(x-1)\arcsin\sqrt{\frac{x}{x+1}}\right)' = x'+\left((x-1)\arcsin\sqrt{\frac{x}{x+1}}\right)' = \\
= 1 + (x-1)'\arcsin\sqrt{\frac{x}{x+1}} + (x-1)\left(\arcsin\sqrt{\frac{x}{x+1}}\right)' =\\
= 1 + \arcsin\sqrt{\frac{x}{x+1}} + (x-1)\cdot\frac{1}{\sqrt{1-\left(\sqrt{\frac{x}{x+1}}\right)^2}}\cdot \left(\sqrt{\frac{x}{x+1}}\right)' =\\
= 1 + \arcsin\sqrt{\frac{x}{x+1}} + \frac{x-1}{\sqrt{1-\frac{x}{x+1}}}\cdot\frac{1}{2}\cdot\frac{1}{\sqrt{\frac{x}{x+1}}}\cdot\left(\frac{x}{x+1}\right)' =\\
= 1 + \arcsin\sqrt{\frac{x}{x+1}} + \frac{x-1}{\sqrt{\frac{x+1-x}{x+1}}}\cdot\frac{\sqrt{x+1}}{2\sqrt{x}}\cdot\frac{x+1-x}{(x+1)^2} =\\
= 1 + \arcsin\sqrt{\frac{x}{x+1}} + \frac{(x-1)\cdot\sqrt{x+1}\cdot\sqrt{x+1}}{\sqrt{1}\cdot 2\sqrt{x}\cdot (x+1)^2} =\\
= 1 + \arcsin\sqrt{\frac{x}{x+1}} + \frac{x-1}{2\sqrt{x}(x+1)} .
\end{multline}\)
Ответ.
\[f'(x) = 1 + \arcsin\sqrt{\frac{x}{x+1}} + \frac{x-1}{2\sqrt{x}(x+1)}.\]
