If `y=log{((1+x)/(1-x))^(1//4)}-(1)/(2)tan^(-1)x," then "(dy)/(dx)=`

Options

A. `(x)/(1-x^(2))`
B. `(x^(2))/(1-x^(4))`
C. `(x)/(1+x^(4))`
D. `(x)/(1-x^(4))`

Answer

To solve the problem, we need to differentiate the function given by:

\[ y = \log\left(\frac{1+x}{1-x}\right)^{\frac{1}{4}} - \frac{1}{2} \tan^{-1}(x) \]

### Step 1: Rewrite the Logarithm
Using the property of logarithms, we can rewrite the equation as:

\[ y = \frac{1}{4} \log\left(\frac{1+x}{1-x}\right) - \frac{1}{2} \tan^{-1}(x) \]

### Step 2: Differentiate the First Term
Now, we differentiate \( y \) with respect to \( x \):

Using the chain rule, the derivative of \( \log(u) \) is \( \frac{1}{u} \cdot \frac{du}{dx} \). Here, \( u = \frac{1+x}{1-x} \).

1. Differentiate \( \frac{1}{4} \log\left(\frac{1+x}{1-x}\right) \):
- \( \frac{d}{dx} \left(\frac{1}{4} \log\left(\frac{1+x}{1-x}\right)\right) = \frac{1}{4} \cdot \frac{1}{\frac{1+x}{1-x}} \cdot \frac{d}{dx}\left(\frac{1+x}{1-x}\right) \)

2. Now, find \( \frac{d}{dx}\left(\frac{1+x}{1-x}\right) \):
- Using the quotient rule: \( \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \)
- Here, \( u = 1+x \) and \( v = 1-x \)
- \( \frac{du}{dx} = 1 \) and \( \frac{dv}{dx} = -1 \)
- Thus,
\[ \frac{d}{dx}\left(\frac{1+x}{1-x}\right) = \frac{(1-x)(1) - (1+x)(-1)}{(1-x)^2} = \frac{(1-x) + (1+x)}{(1-x)^2} = \frac{2}{(1-x)^2} \]

3. Substitute back into the derivative:
\[ \frac{d}{dx}\left(\frac{1}{4} \log\left(\frac{1+x}{1-x}\right)\right) = \frac{1}{4} \cdot \frac{1-x}{1+x} \cdot \frac{2}{(1-x)^2} = \frac{1}{2(1+x)(1-x)} \]

### Step 3: Differentiate the Second Term
Now, differentiate the second term \( -\frac{1}{2} \tan^{-1}(x) \):
\[ \frac{d}{dx}\left(-\frac{1}{2} \tan^{-1}(x)\right) = -\frac{1}{2} \cdot \frac{1}{1+x^2} \]

### Step 4: Combine the Derivatives
Now combine the derivatives from both terms:
\[ \frac{dy}{dx} = \frac{1}{2(1+x)(1-x)} - \frac{1}{2(1+x^2)} \]

### Step 5: Simplify the Expression
To simplify, we can find a common denominator:
The common denominator is \( 2(1+x)(1-x)(1+x^2) \).

1. Rewrite each term:
\[ \frac{1}{2(1+x)(1-x)} = \frac{(1+x^2)}{2(1+x)(1-x)(1+x^2)} \]
\[ -\frac{1}{2(1+x^2)} = -\frac{(1+x)(1-x)}{2(1+x)(1-x)(1+x^2)} \]

2. Combine:
\[ \frac{dy}{dx} = \frac{(1+x^2) - (1+x)(1-x)}{2(1+x)(1-x)(1+x^2)} \]

3. Simplify the numerator:
\[ (1+x^2) - (1+x-x+x^2) = (1+x^2) - (1+x-x+x^2) = (1+x^2) - (1+x^2) = 0 \]

Thus, we find:
\[ \frac{dy}{dx} = \frac{0}{2(1+x)(1-x)(1+x^2)} = 0 \]

### Final Answer
\[ \frac{dy}{dx} = 0 \]