Монгол Бодлогын Сан

Заавар:

Бодолт: $$\begin{align*} f'(x) &= \left(\dfrac{\sqrt {x} }{1 - x\sqrt {x}}:\dfrac{\sqrt {x} + x}{x + \sqrt {x} + 1}\right)'\\ &= \left(\dfrac{\sqrt {x} }{1 - x\sqrt {x}}\cdot\dfrac{x + \sqrt {x} + 1}{\sqrt {x} + x}\right)'\\ &= \left(\dfrac{\sqrt {x} }{(1-\sqrt{x})(1+\sqrt{x}+x)}\cdot\dfrac{x + \sqrt {x} + 1}{\sqrt {x}\cdot (\sqrt{x}+1)}\right)'\\ &=\left(\dfrac{1}{(1-\sqrt{x})(\sqrt{x}+1)}\right)'=\left(\dfrac{1}{1-x}\right)'\\ &=-\dfrac{1}{(1-x)^2}\cdot(-1)=\dfrac{1}{(1-x)^2}\\ f'\left(\dfrac13\right)&=\dfrac{1}{\left(1-\frac13\right)^2}=\dfrac{9}{4}=2.25 \end{align*}$$