If y = $\ln \begin{pmatrix} \frac{\sqrt{1+X}-\sqrt{1-X}}{\sqrt{1+X}+\sqrt{1-X}}\end{pmatrix}$, how will you find $\frac{dy}{dX}$.[Edited]
On
$$\ln(x)=\log_e x$$
Just like any other logarithm, the natural logarithm is the power you need to raise $e$ to get $x$. Because the natural logarithm is so important, we often afford it its own notation (but you do see people write simply $\log x$, without specifying the base, to mean $\ln(x)$).
Michael Hoppe's answer seems to be the best approach for solving the question at hand. Note that he skipped out a middle step. After rewriting the problem as $e^y =$ ..., he squared both sides to give $e^{2y}$ on the LHS, eliminating all of the square root signs on the RHS.
On
First, we can multiply numerator and denominator by $\sqrt{1+x}+\sqrt{1-x}$, obtaining: $$y=\ln(\frac{1}{2}(\sqrt{1+x}+\sqrt{1-x})^2)$$ Remembering the definition of the derivate of $\ln(x)$, we deduce: $$f'(x)=\frac{1}{\frac{1}{2}(\sqrt{1+x}+\sqrt{1-x})^2}=\frac{2}{(\sqrt{1+x}+\sqrt{1-x})^2}$$
On
I think the simplest way to approach the question is to use two theorems:
$\displaystyle\ln\left(\frac{a}{b}\right)=\ln(a)-\ln(b)$. This is one of the Laws of Logarithms, which you should be (or make yourself) familiar with.
The chain rule: $\displaystyle (f\circ g)'(x)=f'(g(x))\cdot g'(x)$. In words, the derivative of a composition can be reduced to a product of the derivative of the outer function and the derivative of the inner function. For example, the derivative of $\ln(\sqrt{x})$ would be $\displaystyle \underbrace{\left(\frac{\mathrm{d}}{\mathrm{d}g}\ln (g)\right)}_{g=\sqrt{x}}\cdot \left(\frac{\mathrm{d}}{\mathrm{d}x}\sqrt{x}\right)=\left(\frac1g\right)\cdot\left(\frac12x^{-1/2}\right)=\frac1{\sqrt{x}}\cdot\frac12x^{-1/2}$.
This should give you all the tools you need to solve the question.
On
Simplify it first by multiplying the top and bottom by $\sqrt{1+X}+\sqrt{1-X}$: $$y=\ln\frac{\sqrt{1+X}-\sqrt{1-X}}{\sqrt{1+X}+\sqrt{1-X}}=\ln\frac{1-\sqrt{1-X^2}}{X}=\\ \ln (1-\sqrt{1-X^2})-\ln X$$ Now take a derivative: $$y'=\frac{1}{1-\sqrt{1-X^2}}\cdot \frac{X}{\sqrt{1-X^2}}-\frac1X$$ Can you simplify it?
Answer:
$$y'=\frac{1}{X\cdot \sqrt{1-X^2}}$$
(This is for the unedited version ...)
We have $$e^{2y}=\frac{1-x-\sqrt{1-x}}{1-x+\sqrt{1-x}}.$$ Now multiply numerator and denominator by ${1-x-\sqrt{1-x}}$, simplify to $$e^{2y}=1+2\frac{\sqrt{1-x}-1}{x}$$ and take the derivative $$ e^{2y}\cdot2 y'=\left(1+2\frac{\sqrt{1-x}-1}{x}\right)'. $$