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JEE Main 2021
Differentiation
Differentiation
Easy

Question

For x > 1, if (2x) 2y = 4e 2x-2y , then (1 + log e 2x) 2 dydx{{dy} \over {dx}} is equal to :

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Solution

(2x) 2y = 4e 2x-2y 2y\ell n2x = \ell n4 + 2x - 2y y = x+n21+n2x{{x + \ell n2} \over {1 + \ell n2x}} y ' = (1+n2x)(x+n2)1x(1+n2x)2{{\left( {1 + \ell n2x} \right) - \left( {x + \ell n2} \right){1 \over x}} \over {{{\left( {1 + \ell n2x} \right)}^2}}} y '(1+n2x)2=[xn2xn2x]{\left( {1 + \ell n2x} \right)^2} = \left[ {{{x\ell n2x - \ell n2} \over x}} \right]

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