જો y=tan ^{-1}left(frac{log left(frac{e}{x^2} | vedclass

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(A) આપેલ છે કે $y=	an ^{-1}\left(\frac{\log \left(\frac{e}{x^2}ight)}{\log \left(ex^2ight)}ight)+	an ^{-1}\left(\frac{4+2 \log x}{1-8 \log x}\

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(A) આપેલ છે કે $y=	an ^{-1}\left(\frac{\log \left(\frac{e}{x^2}ight)}{\log \left(ex^2ight)}ight)+	an ^{-1}\left(\frac{4+2 \log x}{1-8 \log x}ight)$.લઘુગણકના ગુણધર્મોનો ઉપયોગ કરતા,$\log(\frac{e}{x^2}) = \log e - \log x^2 = 1 - 2\log x$ અને $\log(ex^2) = \log e + \log x^2 = 1 + 2\log x$.તેથી,$y = 	an^{-1}\left(\frac{1-2\log x}{1+2\log x}ight) + 	an^{-1}\left(\frac{4+2\log x}{1-8\log x}ight)$.$	an^{-1} A - 	an^{-1} B = 	an^{-1}\left(\frac{A-B}{1+AB}ight)$ સૂત્રનો ઉપયોગ કરતા,$	an^{-1}(1) - 	an^{-1}(2\log x) = 	an^{-1}\left(\frac{1-2\log x}{1+2\log x}ight)$.તે જ રીતે,$	an^{-1}(4) + 	an^{-1}(2\log x) = 	an^{-1}\left(\frac{4+2\log x}{1-8\log x}ight)$.આ કિંમતો મૂકતા,$y = 	an^{-1}(1) - 	an^{-1}(2\log x) + 	an^{-1}(4) + 	an^{-1}(2\log x)$.$y = 	an^{-1}(1) + 	an^{-1}(4)$.અહીં $y$ અચળ હોવાથી,$\frac{dy}{dx} = 0$ થાય.

જો $y=\tan ^{-1}\left(\frac{\log \left(\frac{e}{x^2}\right)}{\log \left(e x^2\right)}\right)+\tan ^{-1}\left(\frac{4+2 \log x}{1-8 \log x}\right)$ હોય,તો $\frac{d y}{d x}$ શોધો.

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