If `y=cos^(-1)((x^(2n)-1)/(x^(2n)+1)))," then "(1+x^(2n))y_(1)=`

Options

A. `-2nx^(n)`
B. `2nx^(n)`
C. `-2nx^(n-1)`
D. `2nx^(n-1)`

Answer

To solve the given problem, we need to differentiate the function \( y = \cos^{-1} \left( \frac{x^{2n} - 1}{x^{2n} + 1} \right) \) and then find \( (1 + x^{2n}) y' \).

### Step 1: Differentiate \( y \)

Using the chain rule, the derivative of \( y \) with respect to \( x \) is given by:

\[ y' = \frac{d}{dx} \left( \cos^{-1}(u) \right) \quad \text{where } u = \frac{x^{2n} - 1}{x^{2n} + 1} \]

The derivative of \( \cos^{-1}(u) \) is:

\[ \frac{dy}{du} = -\frac{1}{\sqrt{1 - u^2}} \]

Now we need to find \( \frac{du}{dx} \):

\[ u = \frac{x^{2n} - 1}{x^{2n} + 1} \]

Using the quotient rule:

\[ \frac{du}{dx} = \frac{(x^{2n} + 1)(2nx^{2n-1}) - (x^{2n} - 1)(2nx^{2n-1})}{(x^{2n} + 1)^2} \]

Simplifying the numerator:

\[ = \frac{(2nx^{2n-1})(x^{2n} + 1 - x^{2n} + 1)}{(x^{2n} + 1)^2} = \frac{(2nx^{2n-1})(2)}{(x^{2n} + 1)^2} = \frac{4nx^{2n-1}}{(x^{2n} + 1)^2} \]

Now substituting \( u \) and \( \frac{du}{dx} \) back into the derivative of \( y \):

\[ y' = -\frac{1}{\sqrt{1 - u^2}} \cdot \frac{du}{dx} = -\frac{1}{\sqrt{1 - \left( \frac{x^{2n} - 1}{x^{2n} + 1} \right)^2}} \cdot \frac{4nx^{2n-1}}{(x^{2n} + 1)^2} \]

### Step 2: Simplify \( 1 - u^2 \)

Now we need to simplify \( 1 - u^2 \):

\[ 1 - u^2 = 1 - \left( \frac{x^{2n} - 1}{x^{2n} + 1} \right)^2 = \frac{(x^{2n} + 1)^2 - (x^{2n} - 1)^2}{(x^{2n} + 1)^2} \]

Calculating the numerator:

\[ = (x^{2n} + 1)^2 - (x^{2n} - 1)^2 = (x^{4n} + 2x^{2n} + 1) - (x^{4n} - 2x^{2n} + 1) = 4x^{2n} \]

Thus,

\[ 1 - u^2 = \frac{4x^{2n}}{(x^{2n} + 1)^2} \]

### Step 3: Substitute back into \( y' \)

Now substituting back into the expression for \( y' \):

\[ y' = -\frac{1}{\sqrt{\frac{4x^{2n}}{(x^{2n} + 1)^2}}} \cdot \frac{4nx^{2n-1}}{(x^{2n} + 1)^2} \]

This simplifies to:

\[ y' = -\frac{(x^{2n} + 1)}{2\sqrt{x^{2n}}} \cdot \frac{4nx^{2n-1}}{(x^{2n} + 1)^2} = -\frac{2n \cdot 2x^{2n-1}}{(x^{2n} + 1)\sqrt{x^{2n}}} \]

### Step 4: Find \( (1 + x^{2n})y' \)

Now we need to find \( (1 + x^{2n})y' \):

\[ (1 + x^{2n})y' = (1 + x^{2n}) \left( -\frac{2n \cdot 2x^{2n-1}}{(x^{2n} + 1)\sqrt{x^{2n}}} \right) \]

This simplifies to:

\[ = -\frac{4n \cdot x^{2n-1}(1 + x^{2n})}{(x^{2n} + 1)\sqrt{x^{2n}}} = -\frac{4n \cdot x^{2n-1}}{\sqrt{x^{2n}}} \]

### Final Result

Thus, the final result is:

\[ (1 + x^{2n})y' = -\frac{4n \cdot x^{n-1}}{1} \]