For x1, if (2x)^2y=4e^2x-2y, then (1+_e2x)^2dydx is equal to:

Question

For $x>1$, if $(2x)^{2y}=4e^{2x-2y}$, then $(1+\log_e2x)^2\dfrac{dy}{dx}$ is equal to:

Accepted Answer

Taking log: $2y\ln(2x)=\ln4+2x-2y$.

Differentiating: $\dfrac{2y}{x}+2\ln(2x)\dfrac{dy}{dx}=2-2\dfrac{dy}{dx}$.

$$(1+\ln2x)\frac{dy}{dx}=1-\frac yx=\frac{x-y}{x}$$

Also from the log equation, $y\ln2x=\ln2+x-y\Rightarrow y=\dfrac{\ln2+x}{1+\ln2x}$.

$$(1+\ln2x)^2\frac{dy}{dx}=(1+\ln2x)\left(x-\frac{\ln2+x}{1+\ln2x}\right)=x(1+\ln2x)-(\ln2+x)=x\ln2x-\ln2$$

Solutions · 1