If `y = tan ^(-1) ((2x)/( 1+15x^(2)) ) , and (dy)/(dx) = ( lambda)/( 1+25x^(2)) -( mu)/( 1+9x^(2)) ," then " lambda + mu ` equal

Options

A. `8`
B. ` 2`
C. ` 10 `
D. ` 34`

Answer

To solve the problem, we start with the given equation:

\[ y = \tan^{-1}\left(\frac{2x}{1 + 15x^2}\right) \]

We will use the identity for the tangent inverse function:

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

### Step 1: Rewrite the expression

We can rewrite \( \frac{2x}{1 + 15x^2} \) in a form that allows us to apply the identity. Notice that we can express \( 2x \) as \( 5x - 3x \) and \( 15x^2 \) as \( 3x \cdot 5x \):

\[ y = \tan^{-1}\left(\frac{5x - 3x}{1 + 3x \cdot 5x}\right) \]

### Step 2: Apply the identity

Using the identity, we can separate the terms:

\[ y = \tan^{-1}(5x) - \tan^{-1}(3x) \]

### Step 3: Differentiate \( y \)

Now, we differentiate \( y \) with respect to \( x \):

\[ \frac{dy}{dx} = \frac{d}{dx}[\tan^{-1}(5x)] - \frac{d}{dx}[\tan^{-1}(3x)] \]

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

\[ \frac{dy}{dx} = \frac{1}{1 + (5x)^2} \cdot 5 - \frac{1}{1 + (3x)^2} \cdot 3 \]

### Step 4: Simplify the derivatives

Calculating the derivatives:

\[ \frac{dy}{dx} = \frac{5}{1 + 25x^2} - \frac{3}{1 + 9x^2} \]

### Step 5: Compare with the given expression

We are given that:

\[ \frac{dy}{dx} = \frac{\lambda}{1 + 25x^2} - \frac{\mu}{1 + 9x^2} \]

By comparing both expressions, we can identify:

\[ \lambda = 5 \quad \text{and} \quad \mu = 3 \]

### Step 6: Find \( \lambda + \mu \)

Now, we can find \( \lambda + \mu \):

\[ \lambda + \mu = 5 + 3 = 8 \]

### Final Answer

Thus, the value of \( \lambda + \mu \) is:

\[ \boxed{8} \]