Question

If $y={\tan }^{-1}\frac{1}{x^2+x+1}+{\tan }^{-1}\frac{1}{x^2+3x+3}+{\tan }^{-1}\frac{1}{x^2+5x+7}+.....$ to n terms, then

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

Answer 1

The correct option is A $\frac{dy}{dx}=\frac{1}{1+(x+n{)}^2}-\frac{1}{1+x^2}$

$y={\tan }^{-1}\frac{1}{x^2+x+1}+{\tan }^{-1}\frac{1}{x^2+3x+3}+{\tan }^{-1}\frac{1}{x^2+5x+7}+$ .... to n terms

$={\tan }^{-1}\left\{\frac{1}{1+x(x+1)}\right\}+{\tan }^{-1}\left\{\frac{1}{1+(x+1)(x+2)}\right\}+{\tan }^{-1}\left\{\frac{1}{1+(x+2)(x+3)}\right\}+.....+{\tan }^{-1}\left\{\frac{1}{1+(x+n-1)\{x+n\}}\right\}$

$={\tan }^{-1}\left\{\frac{(x+1)-x}{1+(x+1)x}\right\}+{\tan }^{-1}\left\{\frac{(x+2)-(x+1)}{1+(x+2)(x+1)}\right\}+{\tan }^{-1}\left\{\frac{(x+3)-(x+2)}{1+(x+3)(x+2)}\right\}+.....+{\tan }^{-1}\left\{\frac{(x+n)-(x+n-1)}{1+(x+n)(x+n-1)}\right\}$

$\therefore y=\left\{{\tan }^{-1}\right(x+1)-{\tan }^{-1}(x\left)\right\}+\left\{{\tan }^{-1}\right(x+2)-{\tan }^{-1}(x+1\left)\right\}+\left\{{\tan }^{-1}\right(x+3)-{\tan }^{-1}(x+2\left)\right\}+.....$

$+\left\{{\tan }^{-1}\right(x+n)-{\tan }^{-1}(x+n-1\left)\right\}$

$y={\tan }^{-1}(x+n)-{\tan }^{-1}\left(x\right)$

On differentiating both the sides w.r.t. x, we get

$\frac{dy}{dx}=\frac{1}{1+(x+n{)}^2}-\frac{1}{1+x^2}$