For x 1, if (2x)2y = 4e2x-2y, then (1 + loge 2x)2 [dy dx] is equal to :

Question

For x > 1,  if  (2x)2y = 4e2x-2y,

then  (1 + loge 2x)2 $${{dy} \over {dx}}$$ is equal to :

Accepted Answer

(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 '$${\left( {1 + \ell n2x} \right)^2} = \left[ {{{x\ell n2x - \ell n2} \over x}} \right]$$

Solutions · 1