If y = tan^{-1}left(frac{1}{x^2 + x + 1}right | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The general term of the series is $T_r = 	an^{-1}\left(\frac{1}{x^2 + (2r-1)x + (r^2-r+1)}ight)$.We can rewrite the argument as $\frac{(x+r) -

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) The general term of the series is $T_r = 	an^{-1}\left(\frac{1}{x^2 + (2r-1)x + (r^2-r+1)}ight)$.We can rewrite the argument as $\frac{(x+r) - (x+r-1)}{1 + (x+r)(x+r-1)}$.Using the formula $	an^{-1}(A) - 	an^{-1}(B) = 	an^{-1}\left(\frac{A-B}{1+AB}ight)$,we get $T_r = 	an^{-1}(x+r) - 	an^{-1}(x+r-1)$.Summing up to $n$ terms:$y = \sum_{r=1}^{n} (	an^{-1}(x+r) - 	an^{-1}(x+r-1)) = 	an^{-1}(x+n) - 	an^{-1}(x)$.Differentiating with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx}(	an^{-1}(x+n)) - \frac{d}{dx}(	an^{-1}(x)) = \frac{1}{1 + (x+n)^2} - \frac{1}{1 + x^2}$.

If $y = \tan^{-1}\left(\frac{1}{x^2 + x + 1}\right) + \tan^{-1}\left(\frac{1}{x^2 + 3x + 3}\right) + \tan^{-1}\left(\frac{1}{x^2 + 5x + 7}\right) + \dots$ up to $n$ terms,then $\frac{dy}{dx}$ is equal to

Similar Questions

If $y = \tan^{-1} \sqrt{\frac{a - x}{a + x}}$,then $\frac{dy}{dx} = $

$\frac{d}{d x} \tan ^{-1}\left[\frac{\sqrt{1+\sin x}-\sqrt{1-\sin x}}{\sqrt{1+\sin x}+\sqrt{1-\sin x}}\right]$ is equal to

If $\sqrt{1 - x^6} + \sqrt{1 - y^6} = a^3(x^3 - y^3)$,then $\frac{dy}{dx} = $

Derivative of $\tan ^{-1}\left(\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}}\right)$ with respect to $\cos ^{-1} x^2$ is

If $f(x) = \sin^{-1}\left(\frac{2 \cdot 3^x}{1 + 9^x}\right)$,then $f'(-\frac{1}{2})$ equals

Vedclass Test Series

Exam Paper Generator

Online Exam Module