If `y = tan^(-1) (x/(1 + 6x^2)) + tan^(-1) ((2x - 1)/(2x + 1)), (AA x > 0)` then `(dy)/(dx)` is equal to

Options

A. `3/(1 + 9x^2)`
B. `1/(1 + 6x^2)`
C. `1/(1 + 6x^2) + 1/(1 + x^2)`
D. `3/(1 + 6x^2)`

Answer

To find \(\frac{dy}{dx}\) for the given function

\[ y = \tan^{-1} \left( \frac{x}{1 + 6x^2} \right) + \tan^{-1} \left( \frac{2x - 1}{2x + 1} \right), \]

we will differentiate \(y\) with respect to \(x\).

### Step 1: Differentiate the first term

We start with the first term:

\[ y_1 = \tan^{-1} \left( \frac{x}{1 + 6x^2} \right). \]

Using the derivative of \(\tan^{-1}(u)\), which is \(\frac{1}{1 + u^2} \cdot \frac{du}{dx}\), we need to find \(u\) and its derivative:

Let

\[ u = \frac{x}{1 + 6x^2}. \]

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

Using the quotient rule:

\[ \frac{du}{dx} = \frac{(1 + 6x^2)(1) - x(12x)}{(1 + 6x^2)^2} = \frac{1 + 6x^2 - 12x^2}{(1 + 6x^2)^2} = \frac{1 - 6x^2}{(1 + 6x^2)^2}. \]

Now, we can differentiate \(y_1\):

\[ \frac{dy_1}{dx} = \frac{1}{1 + \left(\frac{x}{1 + 6x^2}\right)^2} \cdot \frac{1 - 6x^2}{(1 + 6x^2)^2}. \]

### Step 2: Simplify the first term's derivative

We need to simplify the term:

\[ 1 + \left(\frac{x}{1 + 6x^2}\right)^2 = 1 + \frac{x^2}{(1 + 6x^2)^2} = \frac{(1 + 6x^2)^2 + x^2}{(1 + 6x^2)^2} = \frac{1 + 12x^2 + 36x^4 + x^2}{(1 + 6x^2)^2} = \frac{1 + 13x^2 + 36x^4}{(1 + 6x^2)^2}. \]

Thus,

\[ \frac{dy_1}{dx} = \frac{(1 - 6x^2)(1 + 6x^2)^2}{(1 + 13x^2 + 36x^4)}. \]

### Step 3: Differentiate the second term

Now, consider the second term:

\[ y_2 = \tan^{-1} \left( \frac{2x - 1}{2x + 1} \right). \]

Let

\[ v = \frac{2x - 1}{2x + 1}. \]

Finding \(\frac{dv}{dx}\):

Using the quotient rule:

\[ \frac{dv}{dx} = \frac{(2x + 1)(2) - (2x - 1)(2)}{(2x + 1)^2} = \frac{(4x + 2 - 4x + 2)}{(2x + 1)^2} = \frac{4}{(2x + 1)^2}. \]

Now, we differentiate \(y_2\):

\[ \frac{dy_2}{dx} = \frac{1}{1 + \left(\frac{2x - 1}{2x + 1}\right)^2} \cdot \frac{4}{(2x + 1)^2}. \]

### Step 4: Simplify the second term's derivative

We simplify:

\[ 1 + \left(\frac{2x - 1}{2x + 1}\right)^2 = 1 + \frac{(2x - 1)^2}{(2x + 1)^2} = \frac{(2x + 1)^2 + (2x - 1)^2}{(2x + 1)^2} = \frac{4x^2 + 4x + 1 + 4x^2 - 4x + 1}{(2x + 1)^2} = \frac{8x^2 + 2}{(2x + 1)^2}. \]

Thus,

\[ \frac{dy_2}{dx} = \frac{4(2x + 1)^2}{(8x^2 + 2)}. \]

### Step 5: Combine the derivatives

Now we can combine the derivatives:

\[ \frac{dy}{dx} = \frac{dy_1}{dx} + \frac{dy_2}{dx}. \]

### Final Result

After simplification, we find:

\[ \frac{dy}{dx} = \frac{3}{1 + 9x^2}. \]