y = operatorname{Tan}^{-1}left(frac{x}{1+2x^2 | vedclass

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

Question

(B) આપણે જાણીએ છીએ કે $\operatorname{Tan}^{-1}(a) - \operatorname{Tan}^{-1}(b) = \operatorname{Tan}^{-1}\left(\frac{a-b}{1+ab}\right)$.આપણે પદોને આ રી

Vedclass · Accepted Answer

$\frac{3}{9x^2+1} - \frac{1}{x^2+1}$

Vedclass · Answer

(B) આપણે જાણીએ છીએ કે $\operatorname{Tan}^{-1}(a) - \operatorname{Tan}^{-1}(b) = \operatorname{Tan}^{-1}\left(\frac{a-b}{1+ab}\right)$.આપણે પદોને આ રીતે ફરીથી લખી શકીએ:$\operatorname{Tan}^{-1}\left(\frac{x}{1+2x^2}\right) = \operatorname{Tan}^{-1}\left(\frac{2x-x}{1+(2x)(x)}\right) = \operatorname{Tan}^{-1}(2x) - \operatorname{Tan}^{-1}(x)$.$\operatorname{Tan}^{-1}\left(\frac{x}{1+6x^2}\right) = \operatorname{Tan}^{-1}\left(\frac{3x-2x}{1+(3x)(2x)}\right) = \operatorname{Tan}^{-1}(3x) - \operatorname{Tan}^{-1}(2x)$.આ કિંમતોને $y$ ના સમીકરણમાં મૂકતા:$y = (\operatorname{Tan}^{-1}(2x) - \operatorname{Tan}^{-1}(x)) + (\operatorname{Tan}^{-1}(3x) - \operatorname{Tan}^{-1}(2x)) = \operatorname{Tan}^{-1}(3x) - \operatorname{Tan}^{-1}(x)$.$x$ ની સાપેક્ષમાં વિકલન કરતા:$\frac{dy}{dx} = \frac{d}{dx}(\operatorname{Tan}^{-1}(3x)) - \frac{d}{dx}(\operatorname{Tan}^{-1}(x)) = \frac{3}{1+(3x)^2} - \frac{1}{1+x^2} = \frac{3}{9x^2+1} - \frac{1}{x^2+1}$.

$y = \operatorname{Tan}^{-1}\left(\frac{x}{1+2x^2}\right) + \operatorname{Tan}^{-1}\left(\frac{x}{1+6x^2}\right)$ હોય,તો $\frac{dy}{dx} = $

Similar Questions

જો $u=\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}\right)$ અને $v=\tan ^{-1}\left(\frac{2 x \sqrt{1-x^{2}}}{1-2 x^{2}}\right)$ હોય,તો $x=0$ આગળ $\frac{d u}{d v}$ ની કિંમત શોધો.

$y = \sin^{-1}\left(\frac{\sqrt{1+x} - \sqrt{1-x}}{2}\right)$ નું વિકલન શું થાય?

જો $y=\tan ^{-1}\left(\frac{2+3 x}{3-2 x}\right)+\tan ^{-1}\left(\frac{4 x}{1+5 x^2}\right)$ હોય,તો $\frac{d y}{d x}=$

$\frac{d}{dx} \left[ \tan^{-1} \left( \frac{\sqrt{1 + x^2} + \sqrt{1 - x^2}}{\sqrt{1 + x^2} - \sqrt{1 - x^2}} \right) \right] = $

જો $f(x) = \tan^{-1}\left(\frac{1}{\sin^2 x + \sin x + 1}\right) + \tan^{-1}\left(\frac{1}{\sin^2 x + 3\sin x + 3}\right) + \tan^{-1}\left(\frac{1}{\sin^2 x + 5\sin x + 7}\right) + \dots$ $10$ પદો સુધી હોય,તો $f'(0) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module