For $x1$, if $(2 x)^{2 y}=4 \mathrm{e}^{2 x-2 y}$, then $(1+\log 2 x)^2 \frac{d y}{d x}$ is equal to

Question

For $x>1$, if $(2 x)^{2 y}=4 \mathrm{e}^{2 x-2 y}$, then $(1+\log 2 x)^2 \frac{d y}{d x}$ is equal to

Aicharya · Accepted Answer

$(2 x)^{2 y}=4 \mathrm{e}^{2 x-2 y}$ Taking log on both sides, we get $\begin{aligned} & 2 y \log 2 x=\log \left(4 \mathrm{e}^{2 x-2 y}ight) \ & \Rightarrow 2 y \log 2 x=\log 4+\log \mathrm{e}^{2 x-2 y} \ & \Rightarrow 2 y \log 2 x=\log 4+2 x-2 y \ & \Rightarrow 2 y \log 2 x+2 y=\log 4+2 x \ & \Rightarrow 2(y \log 2 x+y)=2 \log 2+2 x \ & \Rightarrow y \log 2 x+y=\log 2+x \ & \Rightarrow y(1+\log 2 x)=x+\log 2 \ & \Rightarrow y=\frac{x+\log 2}{1+\log 2 x} \end{aligned}$ Differentiating w.r.t. $x$, we get $\begin{gathered} \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{(1+\log 2 x)(1+0)-(x+\log 2)\left(\frac{1}{2 x}ight) \cdot 2}{(1+\log 2 x)^2} \ \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{(1+\log 2 x)-\frac{1}{x}(x+\log 2)}{(1+\log 2 x)^2} \ \Rightarrow(1+\log 2 x)^2 \frac{\mathrm{d} y}{\mathrm{~d} x}=1+\log 2 x-1-\frac{\log 2}{x} \ \Rightarrow(1+\log 2 x)^2 \frac{\mathrm{d} y}{\mathrm{~d} x}=\log 2 x-\frac{\log 2}{x} \ =\frac{x \log 2 x-\log 2}{x} \end{gathered}$

For $x>1$, if $(2 x)^{2 y}=4 \mathrm{e}^{2 x-2 y}$, then $(1+\log 2 x)^2 \frac{d y}{d x}$ is equal to

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