1.For x>1x > 1x>1, if (2x)2y=4e2x−2y(2x)^{2y} = 4e^{2x-2y}(2x)2y=4e2x−2y, then (1+log2x)2dydx(1 + \log 2x)^2 \frac{dy}{dx}(1+log2x)2dxdy is equal toa.xlog2x+log2x\frac{x \log 2x + \log 2}{x}xxlog2x+log2b.xlog2x−log2x\frac{x \log 2x - \log 2}{x}xxlog2x−log2c.xlog2xx \log 2xxlog2xLogin to continueOnly logged in users canattempt or see the solution.
