JEE Main 2021DifferentiationDifferentiationEasyQuestionIf x=ey+ey+ey+.....∞x = {e^{y + {e^y} + {e^{y + .....\infty }}}}x=ey+ey+ey+.....∞ , x>0,x > 0,x>0, then dydx{{{dy} \over {dx}}}dxdy isOptionsA1+xx{{1 + x} \over x}x1+xB1x{1 \over x}x1C1−xx{{1 - x} \over x}x1−xDx1+x{x \over {1 + x}}1+xxCheck AnswerHide SolutionSolutionx=ey+ey+.....∞ ⇒x=ey+x.x = {e^{y + {e^{y + .....\infty }}}}\,\, \Rightarrow x = {e^{y + x}}.x=ey+ey+.....∞⇒x=ey+x. Taking log. log x=y+x\log \,\,x = y + xlogx=y+x ⇒1x=dydx+1 \Rightarrow {1 \over x} = {{dy} \over {dx}} + 1⇒x1=dxdy+1 ⇒dydx=1x−1=1−xx \Rightarrow {{dy} \over {dx}} = {1 \over x} - 1 = {{1 - x} \over x}⇒dxdy=x1−1=x1−x