First slide

Derivatives

Question

$If y = \tan^{- 1} \frac{1}{1 + x + x^{2}} + \tan^{- 1} \frac{1}{x^{2} + 3 x + 3} + \tan^{- 1} \frac{1}{x^{2} + 5 x + 7} + \dots . . up to n terms then y^{'} (0) =$

Moderate

A

$- \frac{1}{(n^{2} + 1)}$
B

$- \frac{n^{2}}{(n^{2} + 1)}$
C

$\frac{n^{2}}{(n^{2} + 1)}$
D

$\frac{n^{2}}{(1 - n^{2})}$

Solution

$\begin{array}{l} y = \tan^{- 1} \frac{1}{1 + x + x^{2}} + \tan^{- 1} \frac{1}{x^{2} + 3 x + 3} + \cdot \cdot \cdot + u p t o n t e r m s \\ = \tan^{- 1} \frac{(x + 1) - x}{1 + x (x + 1)} + \tan^{- 1} \frac{(x + 2) - (x + 1)}{1 + (x + 1) (x + 2)} + \cdot \cdot \cdot n t e r m s \\ \tan^{- 1} (x + 1) - \tan^{- 1} x + \tan^{- 1} (x + 2) - \tan^{- 1} (x + 1) + \cdot \cdot \cdot + \tan^{- 1} (x + n) - \tan^{- 1} (x + (n - 1)) \\ = \tan^{- 1} (x + n) - \tan^{- 1} x \\ y (x) = \frac{1}{1 + {(x + n)}^{2}} - \frac{1}{1 + x^{2}} \Rightarrow y' (0) = \frac{1}{1 + n^{2}} - 1 = \frac{- n^{2}}{1 + n^{2}} \end{array}$

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