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 the derivative \( \frac{dy}{dx} \) of the function

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

we will use the properties of the inverse tangent function and the chain rule for differentiation.

### Step 1: Simplify the Expression

We can use the formula for the difference of two inverse tangents:

\[ \tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left( \frac{a + b}{1 - ab} \right), \]

if \( ab < 1 \).

However, in this case, we will differentiate each term separately.

### Step 2: Differentiate the First Term

The first term is

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

Using the derivative of \( \tan^{-1}(u) \):

\[ \frac{d}{dx} \tan^{-1}(u) = \frac{1}{1 + u^2} \cdot \frac{du}{dx}, \]

we set \( u = \frac{x}{1 + 6x^2} \).

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

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

Thus,

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

### Step 3: Simplify the Derivative of the First Term

Now, we simplify \( 1 + \left( \frac{x}{1 + 6x^2} \right)^2 \):

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

Thus,

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

### Step 4: Differentiate the Second Term

Now, we differentiate the second term:

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

Let \( v = \frac{2x - 1}{2x + 1} \).

Using the same derivative formula:

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

Thus,

\[ \frac{dy_2}{dx} = \frac{1}{1 + v^2} \cdot \frac{dv}{dx}. \]

Now, calculate \( 1 + v^2 \):

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

Thus,

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

### Step 5: Combine the Derivatives

Now, we can combine the derivatives:

\[ \frac{dy}{dx} = \frac{dy_1}{dx} + \frac{dy_2}{dx} = \frac{1 - 6x^2}{1 + 13x^2 + 36x^4} + \frac{2}{4x^2 + 1}. \]

### Final Answer

Thus, the final answer for \( \frac{dy}{dx} \) is:

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