Question 28.7.10: Differentiate (i) √1+x/1-x (ii) x/√1-x²...

And, \frac{d t}{d x}=\frac{(1-x) \cdot \frac{d}{d x}(1+x)-(1+x) \cdot \frac{d}{d x}(1-x)}{(1-x)^{2}}\\  \\ \begin{aligned}& =\frac{(1-x) \cdot 1-(1+x)(-1)}{(1-x)^{2}}=\frac{2}{(1-x)^{2}} \\  \\\therefore \quad \frac{d y}{d x} & =\left(\frac{d y}{d t} \times \frac{d t}{d x}\right)=\frac{1}{2 \sqrt{t}} \times \frac{2}{(1-x)^{2}}\\  \\& =\frac{1}{(1-x)^{2} \sqrt{\frac{1+x}{1-x}}}=\frac{\sqrt{1-x}}{(1-x)^{2} \sqrt{1+x}}\\  \\& =\frac{1}{(1-x)^{3 / 2}(1+x)^{1 / 2}} .\end{aligned}