The differential of $f(x)=\sqrt{\dfrac{2-x}{2+x}}$ at $x=0$ and $\delta x=0.15$ is

Sol ${ }^{n}$ : The differential of a function is equal to the derivative of the function times the differtial of the indendent varsiable.

Let $y=f(x)$ be the function, where $x$ is a independent variable and the differential of the independendent vasiable is $\delta x$.

The differential of $f(x)$ is $d y=d(f(x))=f^{\prime}(x) \delta x$ where $f^{\prime}(x)$ denote the derivative of function.

$\begin{array}{l} f(x)=\sqrt{\frac{2-x}{2+x}} \\ \text { Now } \begin{aligned} f^{\prime}(x) & =\frac{-\sqrt{2+x} \cdot \frac{1}{2}(2-x)^{-1 / 2}-\sqrt{2-x} \cdot \frac{1}{2}(2+x)^{-1 / 2}}{(\sqrt{2+x})^{2}} \\ & =-\frac{1}{2} \times \frac{1}{(\sqrt{2+x})^{2}}\left[\frac{\sqrt{2+x}}{\sqrt{2-x}}+\frac{\sqrt{2-x}}{\sqrt{2+x}}\right] \end{aligned} \\ \text { FORMULA } \\ =\frac{-1}{2(2+x)} \cdot \frac{(2+x)+(2-x)}{\sqrt{4-x^{2}}} \\ =-\frac{1 \times 2}{(2+x) \sqrt{4-x^{2}}} \\ \text { If } f(x)=\frac{u}{v} \\ \therefore f^{\prime}(x)=\frac{v v^{\prime}-v v^{\prime}}{v^{2}} \\ u=\sqrt{2-x} \\ \because a^{2}-b^{2} \\ =(a+b)(a-b) \\ \end{array}$

herse

$\begin{array}{l} v=\sqrt{2+x} \\ u=\sqrt{2-x} \end{array}$

$\begin{array}{l} \because a^{2}-b^{2} \\ =(a+b)(a-b) \end{array}$

Now $f^{\prime}(x)$ at 0 i.e $f^{\prime}(0)=-\frac{1}{2}$

So the differential of $f(x)$ at $x=0$ is

$\begin{aligned} f^{\prime}(0) \times \delta x & =-\frac{1}{2} \times 0.15 \quad[\text { Given } \delta x=0.15 \\ & =-0.075 \end{aligned}$