Differentiate

To find: Differentiation of $\frac{\sqrt{a}+\sqrt{x}}{\sqrt{a}-\sqrt{x}}$

Formula used: (i) $\left(\frac{u}{v}\right)^{\prime}=\frac{u^{\prime} v-u v^{\prime}}{v^{2}}$ where $v \neq 0$ (Quotient rule)

(ii) $\frac{d x^{n}}{d x}=n x^{n-1}$

Let us take $u=(\sqrt{a}+\sqrt{x})$ and $v=(\sqrt{a}-\sqrt{x})$

$u^{\prime}=\frac{d u}{d x}=\frac{d(\sqrt{a}+\sqrt{x})}{d x}=\frac{1}{2 \sqrt{x}}$

$v^{\prime}=\frac{d v}{d x}=\frac{d(\sqrt{a}-\sqrt{x})}{d x}=-\frac{1}{2 \sqrt{x}}$

Putting the above obtained values in the formula:-

$\left(\frac{u}{v}\right)^{\prime}=\frac{u^{\prime} v-u v^{\prime}}{v^{2}}$ where $v \neq 0$ (Quotient rule)

$\left[\frac{\sqrt{a}+\sqrt{x}}{\sqrt{a}-\sqrt{x}}\right]^{\prime}=\frac{\frac{1}{2 \sqrt{x}} \times(\sqrt{a}-\sqrt{x})-(\sqrt{a}+\sqrt{x}) \times-\frac{1}{2 \sqrt{x}}}{(\sqrt{a}-\sqrt{x})^{2}}$

$\begin{aligned}=& \frac{\frac{\sqrt{a}}{2 \sqrt{x}}-\frac{1}{2}+\frac{\sqrt{a}}{2 \sqrt{x}}+\frac{1}{2}}{(\sqrt{a}-\sqrt{x})^{2}} \\=& \frac{\sqrt{a}}{\sqrt{x}(\sqrt{a}-\sqrt{x})^{2}} \end{aligned}$

Ans $)=\frac{\sqrt{a}}{\sqrt{x}(\sqrt{a}-\sqrt{x})^{2}}$