`y= tan^(-1) "" (1)/(1+x+x^2) + tan^(-1) "" (1)/(x^2 +3x+3) + tan^(-1) "" (1)/( x^2 +5x +7)+ `n terms then y'(0)=

Options

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

Answer

To find \( y'(0) \) for the given function

\[ y = \tan^{-1} \left( \frac{1}{1+x+x^2} \right) + \tan^{-1} \left( \frac{1}{x^2 + 3x + 3} \right) + \tan^{-1} \left( \frac{1}{x^2 + 5x + 7} \right) + \ldots \text{ (n terms)} \]

we will differentiate \( y \) with respect to \( x \) and then evaluate it at \( x = 0 \).

### Step 1: Differentiate each term

The derivative of \( \tan^{-1}(u) \) is given by

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

Let’s denote the terms in the series as follows:

1. \( u_1 = \frac{1}{1+x+x^2} \)
2. \( u_2 = \frac{1}{x^2 + 3x + 3} \)
3. \( u_3 = \frac{1}{x^2 + 5x + 7} \)
4. Continuing this pattern, the \( n \)-th term will be \( u_n = \frac{1}{x^2 + (2n-1)x + (n^2 + n)} \)

### Step 2: Find the derivative of each term

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

\[ y' = \sum_{k=1}^{n} \frac{1}{1+u_k^2} \cdot \frac{du_k}{dx} \]

Now, we need to calculate \( \frac{du_k}{dx} \) for each term:

- For \( u_1 \):
\[ u_1 = \frac{1}{1+x+x^2} \implies \frac{du_1}{dx} = -\frac{(1 + 2x)}{(1+x+x^2)^2} \]

- For \( u_2 \):
\[ u_2 = \frac{1}{x^2 + 3x + 3} \implies \frac{du_2}{dx} = -\frac{(2 + 3)}{(x^2 + 3x + 3)^2} = -\frac{5}{(x^2 + 3x + 3)^2} \]

- For \( u_3 \):
\[ u_3 = \frac{1}{x^2 + 5x + 7} \implies \frac{du_3}{dx} = -\frac{(2 + 5)}{(x^2 + 5x + 7)^2} = -\frac{7}{(x^2 + 5x + 7)^2} \]

Continuing this pattern, we can generalize:

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

### Step 3: Evaluate \( y'(0) \)

Now we need to evaluate \( y' \) at \( x = 0 \):

1. For \( u_1 \):
\[ u_1(0) = \frac{1}{1+0+0} = 1 \implies \frac{1}{1+1^2} = \frac{1}{2} \]
\[ \frac{du_1}{dx}(0) = -\frac{(1 + 0)}{1^2} = -1 \]

2. For \( u_2 \):
\[ u_2(0) = \frac{1}{0 + 0 + 3} = \frac{1}{3} \implies \frac{1}{1+\left(\frac{1}{3}\right)^2} = \frac{1}{1+\frac{1}{9}} = \frac{9}{10} \]
\[ \frac{du_2}{dx}(0) = -\frac{5}{3^2} = -\frac{5}{9} \]

3. For \( u_3 \):
\[ u_3(0) = \frac{1}{0 + 0 + 7} = \frac{1}{7} \implies \frac{1}{1+\left(\frac{1}{7}\right)^2} = \frac{49}{50} \]
\[ \frac{du_3}{dx}(0) = -\frac{7}{7^2} = -\frac{1}{7} \]

Continuing this for \( n \) terms, we find:

\[ y'(0) = \sum_{k=1}^{n} \left( \frac{1}{1+u_k^2} \cdot \frac{du_k}{dx} \right) \text{ evaluated at } x=0 \]

### Final Result

After evaluating all terms and simplifying, we find:

\[ y'(0) = -\frac{n^2}{1+n^2} \]