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

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(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

$y = \tan^{-1} \left[ \frac{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}}{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}} \right]$ નું $x$ ની સાપેક્ષમાં વિકલન શું થાય?

જો $y = \tan^2 \left( \cos^{-1} \sqrt{\frac{1+x^2}{2}} \right)$ હોય, તો $\frac{dy}{dx} = $

જો $a > b > 0$ અને $x$ લઘુકોણ હોય,તો $\frac{d}{dx} \left[ \cos^{-1} \left( \frac{b - a \cos x}{a - b \cos x} \right) \right] = $

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

$\frac{d}{dx}\left( \tan^{-1} \left( \frac{\cos x}{1 + \sin x} \right) \right) = $

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