If y = sqrt {frac{{1 + x}}{{1 - x}}} , then f | vedclass

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(B) Given $y = \sqrt {\frac{{1 + x}}{{1 - x}}} .$Taking natural logarithm on both sides:$\ln y = \frac{1}{2} \ln(1 + x) - \frac{1}{2} \ln(1 - x).$Diff

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$\frac{1}{{{{(1 + x)}^{1/2}}{{(1 - x)}^{3/2}}}}$

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(B) Given $y = \sqrt {\frac{{1 + x}}{{1 - x}}} .$Taking natural logarithm on both sides:$\ln y = \frac{1}{2} \ln(1 + x) - \frac{1}{2} \ln(1 - x).$Differentiating both sides with respect to $x$:$\frac{1}{y} \frac{{dy}}{{dx}} = \frac{1}{2(1 + x)} - \frac{1}{2} \cdot \frac{1}{(1 - x)} \cdot (-1)$$\frac{1}{y} \frac{{dy}}{{dx}} = \frac{1}{2(1 + x)} + \frac{1}{2(1 - x)}$$\frac{1}{y} \frac{{dy}}{{dx}} = \frac{1}{2} \left[ \frac{(1 - x) + (1 + x)}{(1 + x)(1 - x)} \right]$$\frac{1}{y} \frac{{dy}}{{dx}} = \frac{1}{2} \left[ \frac{2}{1 - {x^2}} \right] = \frac{1}{1 - {x^2}}$$\frac{{dy}}{{dx}} = \frac{y}{1 - {x^2}} = \frac{1}{1 - {x^2}} \cdot \sqrt {\frac{{1 + x}}{{1 - x}}} $$\frac{{dy}}{{dx}} = \frac{1}{{(1 - x)(1 + x)}} \cdot \frac{{{{(1 + x)}^{1/2}}}}{{{{(1 - x)}^{1/2}}}} = \frac{{{{(1 + x)}^{1/2}}}}{{{{(1 + x)}^1}{{(1 - x)}^{3/2}}}} = \frac{1}{{{{(1 + x)}^{1/2}}{{(1 - x)}^{3/2}}}}.$

If $y = \sqrt {\frac{{1 + x}}{{1 - x}}} ,$ then $\frac{{dy}}{{dx}} = $

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