Question

if x^x =y^y ,then dy/Dx is equal to​

Answer 1

Answer:

The answer is $\frac{1+\log x}{1+\log y}$

Step-by-step explanation:

Given $x^x=y^y$

Take log both sides

$\log x^x=\log y^y\Rightarrow x\log x=y\log y$

Now differentiate both sides w.r.t. x

$\frac{d}{dx}(x\log x)=\frac{d}{dx}(y\log y)=\frac{d}{dy}(y\log y)\frac{dy}{dx}\\\Rightarrow \log x(1)+x(\frac{1}{x})=(\log y(1)+y(\frac{1}{y}))\frac{dy}{dx}\\\Rightarrow \log x+1=(\log y+1)\frac{dy}{dx}\\\Rightarrow \frac{dy}{dx}=\frac{1+\log x}{1+\log y}$

Answer 2

Answer:

Solution

$\begin{gathered}\frac{d}{dx}(x\log x)=\frac{d}{dx}(y\log y)=\frac{d}{dy}(y\log y)\frac{dy}{dx}\\\Rightarrow \log x(1)+x(\frac{1}{x})=(\log y(1)+y(\frac{1}{y}))\frac{dy}{dx}\\\Rightarrow \log x+1=(\log y+1)\frac{dy}{dx}\\\Rightarrow \frac{dy}{dx}=\frac{1+\log x}{1+\log y}\end{gathered}$