Question

If $y=\tan^{-1}(\frac{1}{1+x+x^{2}})+\tan^{-1}(\frac{1}{x^{2}+3x+3})+\tan^{-1}(\frac{1}{x^{2}+5x+7})$ then $y^{\prime}(0)$ is}

• A. $9/10$
• B. $1/10$
• C. $-9/10$
• D. $-1/10$

Hint:

In telescoping series for $\tan^{-1}$, only the last and first terms remain.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We simplify the expression for $y$ using the property $\tan^{-1} x - \tan^{-1} y = \tan^{-1}\left( \frac{x-y}{1+xy} \right)$ before differentiating.

Step 2: Key Formula or Approach:
Rewrite each term:
1. $\frac{1}{1+x+x^2} = \frac{(x+1)-x}{1+x(x+1)}$
2. $\frac{1}{x^2+3x+3} = \frac{1}{1+(x^2+3x+2)} = \frac{(x+2)-(x+1)}{1+(x+1)(x+2)}$
3. $\frac{1}{x^2+5x+7} = \frac{1}{1+(x^2+5x+6)} = \frac{(x+3)-(x+2)}{1+(x+2)(x+3)}$

Step 3: Detailed Explanation:
Using the properties:
$y = [\tan^{-1}(x+1) - \tan^{-1} x] + [\tan^{-1}(x+2) - \tan^{-1}(x+1)] + [\tan^{-1}(x+3) - \tan^{-1}(x+2)]$
This is a telescoping sum. Most terms cancel out:
$y = \tan^{-1}(x+3) - \tan^{-1} x$.
Differentiating with respect to $x$:
$y' = \frac{1}{1+(x+3)^2} - \frac{1}{1+x^2}$.
At $x = 0$:
$y'(0) = \frac{1}{1+3^2} - \frac{1}{1+0^2} = \frac{1}{10} - 1 = -\frac{9}{10}$.

Step 4: Final Answer:
The value of $y'(0)$ is $-9/10$.