Question

If \[{{x}^{y}}={{y}^{x}}\] , then find the value of \[\dfrac{dy}{dx}\] .

Answer 1

Hint: First of all, take \[\log \] in LHS and RHS of the expression, \[{{x}^{y}}={{y}^{x}}\] . Now, simplify the expression by using the formula, \[\log {{a}^{b}}=b\log a\] . Differentiate the simplified expression with respect to x using the formula, \[\dfrac{d\left( uv \right)}{dx}=u\dfrac{dv}{dx}+v\dfrac{du}{dx}\] . We know the formula, \[\dfrac{d\left( \log x \right)}{dx}=\dfrac{1}{x}\] . Now, use this formula, to simplify the equation, \[y\dfrac{d\left( \log x \right)}{dx}+\log x\dfrac{dy}{dx}=x\dfrac{d\left( \log y \right)}{dx}+\log y\dfrac{dx}{dx}\] . Then, use chain rule to simplify \[\dfrac{d\left( \log y \right)}{dx}\] . Now, solve it further and get the value of \[\dfrac{dy}{dx}\] .

Complete step by step answer:
According to the question, it is given that the expression is
\[{{x}^{y}}={{y}^{x}}\] ……………………………………..(1)
Now, on taking \[\log \] in LHS and RHS of equation (1), we get
\[\Rightarrow \log {{x}^{y}}=\log {{y}^{x}}\] ………………………………………(2)
We know the formula, \[\log {{a}^{b}}=b\log a\] ………………………………………(3)
Now, from equation (2) and equation (3), we get
\[\Rightarrow \log {{x}^{y}}=\log {{y}^{x}}\]
\[\Rightarrow y\log x=x\log y\] ……………………………………….(4)
Now, on differentiating with respect to x in equation (4), we get
\[\Rightarrow \dfrac{d\left( y\log x \right)}{dx}=\dfrac{d\left( x\log y \right)}{dx}\] ……………………………………………(5)
We know the formula, \[\dfrac{d\left( uv \right)}{dx}=u\dfrac{dv}{dx}+v\dfrac{du}{dx}\] ………………………………………………(6)
Now, on applying the formula shown in equation (6) to simplify equation (5), we get
\[\Rightarrow y\dfrac{d\left( \log x \right)}{dx}+\log x\dfrac{dy}{dx}=x\dfrac{d\left( \log y \right)}{dx}+\log y\dfrac{dx}{dx}\] …………………………………………..(7)
We know the formula, \[\dfrac{d\left( \log x \right)}{dx}=\dfrac{1}{x}\] ……………………………………………………(8)
\[\Rightarrow y.\dfrac{1}{x}+\log x\dfrac{dy}{dx}=x\dfrac{d\left( \log y \right)}{dx}+\log y.1\] ……………………………………….(9)
Now, using chain rule, we can write, \[\dfrac{d\left( \log y \right)}{dx}=\dfrac{d\left( \log y \right)}{dy}\times \dfrac{dy}{dx}\] ………………………………………..(10)
Using equation (10) and transforming equation (9), we get
\[\Rightarrow \dfrac{y}{x}+\log x\dfrac{dy}{dx}=x\dfrac{d\left( \log y \right)}{dy}\times \dfrac{dy}{dx}+\log y\] ………………………………………..(11)
Now, on replacing x by y in equation (8), we get
\[\dfrac{d\left( \log y \right)}{dy}=\dfrac{1}{x}\] …………………………………………….(12)
Using equation (12) for simplifying equation (11), we get
\[\begin{align}
  & \Rightarrow \dfrac{y}{x}+\log x\dfrac{dy}{dx}=x.\dfrac{1}{y}\times \dfrac{dy}{dx}+\log y \\
 & \Rightarrow \dfrac{y}{x}+\log x\dfrac{dy}{dx}=\dfrac{x}{y}\dfrac{dy}{dx}+\log y \\
\end{align}\]
Now, on shifting the \[\log \] terms to the LHS and remaining terms to RHS, we get
\[\Rightarrow \log x\dfrac{dy}{dx}-\dfrac{x}{y}\dfrac{dy}{dx}=\log y-\dfrac{y}{x}\]
\[\Rightarrow \dfrac{dy}{dx}\left( \log x-\dfrac{x}{y} \right)=\left( \log y-\dfrac{y}{x} \right)\] …………………………………….(13)
Now, on dividing by the term \[\left( \log x-\dfrac{x}{y} \right)\] in LHS and RHS of equation (13), we get
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{\left( \log y-\dfrac{y}{x} \right)}{\left( \log x-\dfrac{x}{y} \right)}\] ……………………………………(14)
On simplifying equation (14), we get
\[\begin{align}
  & \Rightarrow \dfrac{dy}{dx}=\dfrac{\left( \dfrac{x\log y-y}{x} \right)}{\left( \dfrac{y\log x-x}{y} \right)} \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{y}{x}\left( \dfrac{x\log y-y}{y\log x-x} \right) \\
\end{align}\]
Therefore, the value of \[\dfrac{dy}{dx}\] for the expression \[{{x}^{y}}={{y}^{x}}\] is equal to \[\dfrac{y}{x}\left( \dfrac{x\log y-y}{y\log x-x} \right)\] .

Note:
 In this question, one might do a silly while differentiating the term \[\log y\] with respect to x, that is, \[\dfrac{d\left( \log y \right)}{dx}\] . One might use the formula, \[\dfrac{d\left( \log x \right)}{dx}=\dfrac{1}{x}\] and write it as \[\dfrac{1}{y}\] . This is wrong because here, we can’t use the formula \[\dfrac{d\left( \log x \right)}{dx}=\dfrac{1}{x}\] directly. For using this formula, we must have the same variable in term in numerator and denominator as well. But here, differentiating the term \[\log y\] with respect to x, that is, \[\dfrac{d\left( \log y \right)}{dx}\] , we don’t have same variable in numerator and denominator. Therefore, first transform the term \[\dfrac{d\left( \log y \right)}{dx}\] using the chain rule as \[\dfrac{d\left( \log y \right)}{dy}\times \dfrac{dy}{dx}\] .