JEE MAIN - Mathematics (2019 - 12th January Morning Slot - No. 19)
For x > 1, if (2x)2y = 4e2x$$-$$2y,
then (1 + loge 2x)2 $${{dy} \over {dx}}$$ is equal to :
then (1 + loge 2x)2 $${{dy} \over {dx}}$$ is equal to :
$${{x\,{{\log }_e}2x - {{\log }_e}2} \over x}$$
loge 2x
x loge 2x
$${{x\,{{\log }_e}2x + {{\log }_e}2} \over x}$$
Explanation
(2x)2y = 4e2x-2y
2y$$\ell $$n2x = $$\ell $$n4 + 2x $$-$$ 2y
y = $${{x + \ell n2} \over {1 + \ell n2x}}$$
y ' = $${{\left( {1 + \ell n2x} \right) - \left( {x + \ell n2} \right){1 \over x}} \over {{{\left( {1 + \ell n2x} \right)}^2}}}$$
y '$${\left( {1 + \ell n2x} \right)^2} = \left[ {{{x\ell n2x - \ell n2} \over x}} \right]$$
2y$$\ell $$n2x = $$\ell $$n4 + 2x $$-$$ 2y
y = $${{x + \ell n2} \over {1 + \ell n2x}}$$
y ' = $${{\left( {1 + \ell n2x} \right) - \left( {x + \ell n2} \right){1 \over x}} \over {{{\left( {1 + \ell n2x} \right)}^2}}}$$
y '$${\left( {1 + \ell n2x} \right)^2} = \left[ {{{x\ell n2x - \ell n2} \over x}} \right]$$
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