Монгол Бодлогын Сан

Заавар:$(x^\alpha)'=\alpha x^{\alpha-1}$ болохыг ашигла.

Бодолт:\begin{align*} \left(\dfrac{x+1}{\sqrt{x}}\right)''&=\left(\sqrt{x}+\dfrac{1}{\sqrt{x}}\right)''=\big(x^{\frac12}+x^{-\frac12}\big)''\\ &=\left(\frac12x^{\frac12-1}-\frac12x^{-\frac12-1}\right)^\prime=\left(\frac12x^{-\frac12}-\frac12x^{-\frac32}\right)^\prime\\ &=\frac12\cdot\left(-\dfrac12\right)x^{-\frac12-1}-\frac12\cdot\left(-\dfrac32\right)x^{-\frac32-1}\\ &=-\frac14x^{-\frac32}+\frac34x^{-\frac52}=-\frac{1}{4x\sqrt{x}}+\frac{3}{4x^2\sqrt{x}}=\frac{3-x}{4x^2\sqrt{x}} \end{align*}