If y = 𝐼𝑛 √1+X − √1−X √1+X + √1−X , find dy dX .

 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 \( y = \ln \left( \sqrt{1+X} - \sqrt{1-X} \right) - \ln \left( \sqrt{1+X} + \sqrt{1-X} \right) \).