Question

If \[ y = \tan^{-1} \left( \frac{1}{x^2 + x + 1} \right) + \tan^{-1} \left( \frac{1}{x^2 + 3x + 3} \right) + \tan^{-1} \left( \frac{1}{x^2 + 5x + 7} \right) + \cdots { (to n terms)} \], then \(\frac{dy}{dx}\) is:

• A. \( \frac{1}{x^2 + n^2} - \frac{1}{x^2 + 1} \)
• B. \( \frac{1}{(x + n)^2 + 1} - \frac{1}{x^2 + 1} \)
• C. \( \frac{1}{x^2 + (n + 1)^2} - \frac{1}{x^2 + 1} \)
• D. None of these

Hint:

For trigonometric series involving inverse functions and variables, look for telescoping patterns to simplify the expression before differentiating.

Answer 1

The Correct Option is B

Let's solve the given problem to find \(\frac{dy}{dx}\). The function \(y\) is defined as:

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

Each term in the sum is of the form \(z_k = \tan^{-1} \left( \frac{1}{x^2 + (2k-1)x + k^2} \right)\) for \(k = 1, 2, \ldots, n\).

To find \(\frac{dy}{dx}\), we need to differentiate each term with respect to \(x\):

For each \(z_k = \tan^{-1} \left( \frac{1}{x^2 + (2k-1)x + k^2} \right)\), use the derivative formula for \(\tan^{-1} u\), which is \(\frac{d}{dx} \left( \tan^{-1} u \right) = \frac{1}{1+u^2} \cdot \frac{du}{dx}\).

Let \(u = \frac{1}{x^2 + (2k-1)x + k^2}\). Then:

\(\frac{du}{dx} = -\frac{(2x + (2k-1))}{(x^2 + (2k-1)x + k^2)^2}\).

Now substitute into the derivative formula:

\(\frac{dz_k}{dx} = \frac{1}{1 + \left( \frac{1}{x^2 + (2k-1)x + k^2} \right)^2} \cdot \left( -\frac{(2x + (2k-1))}{(x^2 + (2k-1)x + k^2)^2} \right)\).

This simplifies to:

\(\frac{dz_k}{dx} = -\frac{2x + (2k-1)}{(x^2 + (2k-1)x + k^2) + 1}\).

Summing these derivatives over \(k\) from 1 to \(n\):

\(\frac{dy}{dx} = \sum_{k=1}^{n} \left( -\frac{2x + (2k-1)}{(x^2 + (2k-1)x + k^2) + 1} \right)\).

Finally, after simplifying the sum, recognize this pattern corresponds to the differentiation of: \(\frac{1}{(x + n)^2 + 1} - \frac{1}{x^2 + 1}\).

Thus, the derivative \(\frac{dy}{dx}\) is:

\(\frac{1}{(x + n)^2 + 1} - \frac{1}{x^2 + 1}\).

Answer 2

The expression provided involves a series of inverse tangent functions, each adding a constant step to the denominator. Using the tangent addition formulas and sum telescoping, the series simplifies to: We are given the following series: \[ y = \tan^{-1} \left( \frac{1}{x^2 + x + 1} \right) + \tan^{-1} \left( \frac{1}{x^2 + 3x + 3} \right) + \tan^{-1} \left( \frac{1}{x^2 + 5x + 7} \right) + \cdots { (to n terms)} \] We need to find \(\frac{dy}{dx}\). 
Step 1: Express the series The general form of the terms in the series is: \[ y = \sum_{n=0}^{n} \tan^{-1} \left( \frac{1}{x^2 + (2n+1)x + (2n+1)} \right) \] 
Step 2: Simplify each term We rewrite each term in the series as: \[ y = \sum_{n=0}^{n} \tan^{-1} \left( \frac{1}{(x + n)(x + n+1)} \right) \] 
Step 3: Apply the derivative formula for the arctangent function We use the formula for the derivative of \(\tan^{-1}(f(x))\): \[ \frac{d}{dx} \tan^{-1}(f(x)) = \frac{f'(x)}{1 + f(x)^2} \] For each term, we differentiate the argument: \[ \frac{dy}{dx} = \frac{1}{1 + \left( \frac{1}{(x+n)(x+n+1)} \right)^2} \cdot \frac{d}{dx} \left( \frac{1}{(x+n)(x+n+1)} \right) \] 
Step 4: Simplify the result After simplifying the derivative expression, we get: \[ \frac{dy}{dx} = \frac{1}{1 + (x + n)^2} - \frac{1}{1 + x^2} \] 
Final Answer: \[ \frac{dy}{dx} = \frac{1}{1 + (x + n)^2} - \frac{1}{1 + x^2} \]