Question

If y = 2^logx/2, then dy/dx is :​

Answer 1

Step-by-step explanation:

We have,

$y = \frac{ {2}^{ log(x) } }{2}$

$= > 2y = {2}^{ log_{e}(x) }$

$= > 2y = {x}^{ log_{e}(2) } \: \: \: \: \: (from \: \: base\: \:change \: \: theorem)$

$= > 2y = {x}^{ ln(2) }$

$= > 2 \frac{dy}{dx} = ln(2) ( {x})^{ ln(2) - 1}$

$= > \frac{dy}{dx} = \frac{1}{2}. ln(2) ( {x})^{ ln(2) - ln(e) }$

$= > \frac{dy }{dx} = \frac{1}{2} . ln(2)( {x})^{ ln( \frac{2}{e} ) }$