# If y = πΌπ √1+X − √1−X √1+X + √1−X , find dy dX .

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To find $$\frac{dy}{dX}$$ for $$y = \ln \left( \sqrt{1+X} - \sqrt{1-X} \right) - \ln \left( \sqrt{1+X} + \sqrt{1-X} \right)$$, we'll use the chain rule and the properties of logarithmic differentiation.

Let $$u = \sqrt{1+X} - \sqrt{1-X}$$ and $$v = \sqrt{1+X} + \sqrt{1-X}$$. Then, $$y = \ln(u) - \ln(v)$$.

Now, differentiate $$u$$ and $$v$$ with respect to $$X$$:

$\frac{du}{dX} = \frac{1}{2} \left( \frac{1}{\sqrt{1+X}} - \frac{1}{\sqrt{1-X}} \right)$

$\frac{dv}{dX} = \frac{1}{2} \left( \frac{1}{\sqrt{1+X}} + \frac{1}{\sqrt{1-X}} \right)$

Now, apply the chain rule and the quotient rule to find $$\frac{dy}{dX}$$:

$\frac{dy}{dX} = \frac{1}{u} \cdot \frac{du}{dX} - \frac{1}{v} \cdot \frac{dv}{dX}$

Substitute the expressions for $$\frac{du}{dX}$$ and $$\frac{dv}{dX}$$:

$\frac{dy}{dX} = \frac{1}{\sqrt{1+X} - \sqrt{1-X}} \cdot \frac{1}{2} \left( \frac{1}{\sqrt{1+X}} - \frac{1}{\sqrt{1-X}} \right) - \frac{1}{\sqrt{1+X} + \sqrt{1-X}} \cdot \frac{1}{2} \left( \frac{1}{\sqrt{1+X}} + \frac{1}{\sqrt{1-X}} \right)$

Combine the terms:

$\frac{dy}{dX} = \frac{1}{2} \left( \frac{1}{1+X} - \frac{1}{1-X} \right) - \frac{1}{2} \left( \frac{1}{1+X} + \frac{1}{1-X} \right) \cdot \frac{1}{\sqrt{1+X} + \sqrt{1-X}}$

Combine the fractions:

$\frac{dy}{dX} = \frac{1}{2} \left( \frac{1}{1-X} - \frac{1}{1+X} \right) - \frac{1}{2} \cdot \frac{1}{\sqrt{1+X} + \sqrt{1-X}}$

Multiply the numerator and denominator of the second term by $$\sqrt{1+X} - \sqrt{1-X}$$ to rationalize the denominator:

$\frac{dy}{dX} = \frac{1}{2} \left( \frac{1}{1-X} - \frac{1}{1+X} \right) - \frac{\sqrt{1+X} - \sqrt{1-X}}{2}$

Combine the terms:

$\frac{dy}{dX} = \frac{1}{2} \left( \frac{1}{1-X} - \frac{1}{1+X} \right) - \frac{\sqrt{1+X} - \sqrt{1-X}}{2}$

Now, if you want to simplify further, you can combine the fractions in the first term by finding a common denominator:

$\frac{dy}{dX} = \frac{1}{2} \cdot \frac{2X}{1-X^2} - \frac{\sqrt{1+X} - \sqrt{1-X}}{2}$

Combine the terms:

$\frac{dy}{dX} = \frac{X}{1-X^2} - \frac{\sqrt{1+X} - \sqrt{1-X}}{2}$

This is the derivative $$\frac{dy}{dX}$$ for the given function $$y = \ln \left( \sqrt{1+X} - \sqrt{1-X} \right) - \ln \left( \sqrt{1+X} + \sqrt{1-X} \right)$$.