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

Options

A. `(1)/(1+x^2)`
B. `(1)/(1+(1+x)^2)`
C. 0
D. None

Answer

To solve the problem, we need to differentiate the given function:

Given:
\[ y = \sum_{n=1}^{x} \tan^{-1} \left( \frac{1}{1+n+n^2} \right) \]

### Step 1: Simplify the Argument of the Arctangent
First, we simplify the term inside the summation:
\[ 1 + n + n^2 = n^2 + n + 1 \]
We can rewrite the term:
\[ \tan^{-1} \left( \frac{1}{1+n+n^2} \right) = \tan^{-1} \left( \frac{1}{n^2+n+1} \right) \]

### Step 2: Use the Formula for the Difference of Arctangents
We can use the identity for the difference of arctangents:
\[ \tan^{-1}(a) - \tan^{-1}(b) = \tan^{-1} \left( \frac{a-b}{1+ab} \right) \]
This means we can express the sum as:
\[ y = \sum_{n=1}^{x} \left( \tan^{-1}(n+1) - \tan^{-1}(n) \right) \]

### Step 3: Write Out the Terms
Now, let's write out the terms of the summation:
- For \( n = 1 \): \( \tan^{-1}(2) - \tan^{-1}(1) \)
- For \( n = 2 \): \( \tan^{-1}(3) - \tan^{-1}(2) \)
- For \( n = 3 \): \( \tan^{-1}(4) - \tan^{-1}(3) \)
- ...
- For \( n = x \): \( \tan^{-1}(x+1) - \tan^{-1}(x) \)

### Step 4: Observe the Cancellation
Notice that this is a telescoping series:
\[ y = \tan^{-1}(x+1) - \tan^{-1}(1) \]

### Step 5: Differentiate \( y \)
Now we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = \frac{d}{dx} \left( \tan^{-1}(x+1) - \tan^{-1}(1) \right) \]
Since \( \tan^{-1}(1) \) is a constant, its derivative is 0. Thus, we only need to differentiate \( \tan^{-1}(x+1) \):
\[ \frac{dy}{dx} = \frac{1}{1+(x+1)^2} \cdot \frac{d}{dx}(x+1) = \frac{1}{1+(x+1)^2} \cdot 1 \]
This simplifies to:
\[ \frac{dy}{dx} = \frac{1}{1+(x+1)^2} \]

### Step 6: Final Result
Thus, the final result for the derivative is:
\[ \frac{dy}{dx} = \frac{1}{1 + (x+1)^2} \]