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

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

Aicharya · Accepted Answer

$(2 x)^{2 y}=4 \mathrm{e}^{2 x-2 y}$ Taking ' $\log _{\mathrm{e}}$ ' on both sides, we get $2 y \log _{\mathrm{e}}(2 x)=\log _{\mathrm{e}} 4+(2 x-2 y) \log _{\mathrm{e}} \mathrm{e}$ $	herefore \quad y \log _{\mathrm{e}}(2 x)=\log _{\mathrm{e}} 2+x-y$ Differentiating w.r.t. $x$, we get $\begin{array}{ll} & {\left[\log _{\mathrm{e}} 2 xight] \frac{\mathrm{d} y}{\mathrm{~d} x}+\frac{2 y}{2 x}=0+1-\frac{\mathrm{d} y}{\mathrm{~d} x}} \ & {\left[1+\log _{\mathrm{e}} 2 xight] \frac{\mathrm{d} y}{\mathrm{~d} x}=1-\frac{y}{x}} \ & 	ext { Now, (i) } \Rightarrow y=\frac{\log _{\mathrm{e}} 2+x}{1+\log _{\mathrm{e}} 2 x} \ 	herefore \quad \ldots 	ext { (ii) } & \left(	ext { ii) } \Rightarrow\left(1+\log _{\mathrm{e}} 2 xight) \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{x \log _{\mathrm{e}} 2 x-\log _{\mathrm{e}} 2}{x\left(1+\log _e 2 xight)}ight. \ 	herefore \quad & \left(1+\log _{\mathrm{e}} 2 xight)^2 \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{x \log _{\mathrm{e}} 2 x-\log _{\mathrm{e}} 2}{x} \end{array}$

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

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