If `y=sum_(r=1)^(x) tan^(-1)((1)/(1+r+r^(2)))`, then `(dy)/(dx)` is equal to

Options

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

Answer

To solve the problem, we need to find the derivative of the function defined as:

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

### Step 1: Rewrite the summation
We can rewrite the term inside the summation:

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

### Step 2: Simplify the argument of the arctangent
We can use the identity for the difference of arctangents:

\[ \tan^{-1}(A) - \tan^{-1}(B) = \tan^{-1}\left(\frac{A-B}{1+AB}\right) \]

We can express \(\frac{1}{1+r+r^2}\) in a form that allows us to apply this identity. We can rewrite it as:

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

Thus, we have:

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

### Step 3: Expand the summation
Now we expand the summation:

\[ y = \left( \tan^{-1}(2) - \tan^{-1}(1) \right) + \left( \tan^{-1}(3) - \tan^{-1}(2) \right) + \ldots + \left( \tan^{-1}(x+1) - \tan^{-1}(x) \right) \]

Most terms will cancel out:

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

### Step 4: Evaluate \(\tan^{-1}(1)\)
We know that:

\[ \tan^{-1}(1) = \frac{\pi}{4} \]

Thus, we can write:

\[ y = \tan^{-1}(x+1) - \frac{\pi}{4} \]

### Step 5: Differentiate \(y\) with respect to \(x\)
Now we differentiate \(y\):

\[ \frac{dy}{dx} = \frac{d}{dx} \left( \tan^{-1}(x+1) - \frac{\pi}{4} \right) \]

The derivative of a constant is zero, so we have:

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

Since \(\frac{d}{dx}(x+1) = 1\), we get:

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

### Step 6: Final result
Thus, the final expression for \(\frac{dy}{dx}\) is:

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