The objective is to compute \(\frac{dy}{dx}\) for the given function \(y\):
\[ y = \tan^{-1} \left( \frac{1}{x^2 + x + 1} \right) + \tan^{-1} \left( \frac{1}{x^2 + 3x + 3} \right) + \tan^{-1} \left( \frac{1}{x^2 + 5x + 7} \right) + \cdots \text{{ (to n terms)}} \]
Each term in the series can be represented as \(z_k = \tan^{-1} \left( \frac{1}{x^2 + (2k-1)x + k^2} \right)\), where \(k\) ranges from 1 to \(n\).
To find \(\frac{dy}{dx}\), we differentiate each term \(z_k\) with respect to \(x\).
The derivative of \(\tan^{-1} u\) is \(\frac{1}{1+u^2} \cdot \frac{du}{dx}\). Let \(u = \frac{1}{x^2 + (2k-1)x + k^2}\).
The derivative of \(u\) with respect to \(x\) is:
\(\frac{du}{dx} = -\frac{2x + (2k-1)}{(x^2 + (2k-1)x + k^2)^2}\).
Substituting this into the derivative formula for \(\tan^{-1} u\):
\(\frac{dz_k}{dx} = \frac{1}{1 + \left( \frac{1}{x^2 + (2k-1)x + k^2} \right)^2} \cdot \left( -\frac{2x + (2k-1)}{(x^2 + (2k-1)x + k^2)^2} \right)\).
This expression simplifies to:
\(\frac{dz_k}{dx} = -\frac{2x + (2k-1)}{(x^2 + (2k-1)x + k^2) + 1}\).
The total derivative \(\frac{dy}{dx}\) is the sum of these individual derivatives from \(k=1\) to \(n\):
\(\frac{dy}{dx} = \sum_{k=1}^{n} \left( -\frac{2x + (2k-1)}{(x^2 + (2k-1)x + k^2) + 1} \right)\).
Upon simplification of this sum, it can be observed that this pattern matches the differentiation of \(\frac{1}{(x + n)^2 + 1} - \frac{1}{x^2 + 1}\).
Therefore, the derivative \(\frac{dy}{dx}\) is:
\(\frac{1}{(x + n)^2 + 1} - \frac{1}{x^2 + 1}\).