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

Options

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

Answer

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

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

we will use the differentiation formula for the inverse tangent function. The derivative of \( \tan^{-1}(u) \) with respect to \( x \) is given by:

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

### Step 1: Differentiate \( \tan^{-1}(1 + 2x) \)

Let \( u = 1 + 2x \).

Then,

\[ \frac{du}{dx} = 2. \]

Now, applying the differentiation formula:

\[ \frac{d}{dx} \tan^{-1}(1 + 2x) = \frac{1}{1 + (1 + 2x)^2} \cdot 2. \]

### Step 2: Differentiate \( \tan^{-1}(1 - 2x) \)

Let \( v = 1 - 2x \).

Then,

\[ \frac{dv}{dx} = -2. \]

Now, applying the differentiation formula:

\[ \frac{d}{dx} \tan^{-1}(1 - 2x) = \frac{1}{1 + (1 - 2x)^2} \cdot (-2). \]

### Step 3: Combine the derivatives

Now, we can combine the derivatives from Step 1 and Step 2:

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

### Step 4: Simplify the expression

Now, we need to simplify the expression:

1. Calculate \( (1 + 2x)^2 \) and \( (1 - 2x)^2 \):

\[ (1 + 2x)^2 = 1 + 4x + 4x^2, \]
\[ (1 - 2x)^2 = 1 - 4x + 4x^2. \]

2. Substitute these back into the derivative:

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

3. Factor out the common terms:

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

### Final Result

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

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