If $, y = \ln{\left( \frac{1 + x}{1 - x} \right )} - 2 \tan^{- 1}{x} \text{ } t h e n , \frac{dy}{dx} = ?$

We are given the function:

$y=\ln \left(\frac{1+x}{1-x}\right)-2 \tan ^{-1} x$

We need to differentiate $y$ with respect to $x$.

Step 1: Differentiate the first term

$y_{1}=\ln \left(\frac{1+x}{1-x}\right)$

Using the derivative formula for $\ln f(x)$ :

$\frac{d}{d x} \ln f(x)=\frac{1}{f(x)} \cdot f^{\prime}(x)$

where $f(x)=\frac{1+x}{1-x}$. Differentiate using the quotient rule:

$\begin{array}{c} f^{\prime}(x)=\frac{(1-x)(1)-(1+x)(-1)}{(1-x)^{2}} \\ \quad=\frac{1-x+1+x}{(1-x)^{2}}=\frac{2}{(1-x)^{2}} \end{array}$

So,

$\begin{array}{l} \frac{d}{d x} \ln \left(\frac{1+x}{1-x}\right)=\frac{1}{\frac{1+x}{1-x}} \cdot \frac{2}{(1-x)^{2}} \\ =\frac{(1-x)}{(1+x)} \cdot \frac{2}{(1-x)^{2}} \\ =\frac{2(1-x)}{(1+x)(1-x)^{2}} \\ =\frac{2}{(1-x)(1+x)} \end{array}$

Since $1-x^{2}=(1-x)(1+x)$, we get

$\frac{d}{d x} \ln \left(\frac{1+x}{1-x}\right)=\frac{2}{1-x^{2}}$

Step 2: Differentiate the second term

$y_{2}=-2 \tan ^{-1} x$

Using the derivative formula:

$\frac{d}{d x} \tan ^{-1} x=\frac{1}{1+x^{2}}$

we get:

$\frac{d}{d x}\left(-2 \tan ^{-1} x\right)=-2 \cdot \frac{1}{1+x^{2}}=\frac{-2}{1+x^{2}}$

Step 3: Compute $\frac{d y}{d x}$

$\frac{d y}{d x}=\frac{2}{1-x^{2}}-\frac{2}{1+x^{2}}$

Step 4: Simplify

Taking the LCM:

$\begin{array}{c} \left(1-x^{2}\right)\left(1+x^{2}\right)=1-x^{4} \\ \frac{2\left(1+x^{2}\right)-2\left(1-x^{2}\right)}{1-x^{4}} \\ =\frac{2+2 x^{2}-2+2 x^{2}}{1-x^{4}} \\ =\frac{4 x^{2}}{1-x^{4}} \end{array}$

Thus, the final result is:

$\frac{d y}{d x}=\frac{4 x^{2}}{1-x^{4}}$