Question

If we have an algebraic expression as \[{{x}^{y}}={{y}^{x}}\] then, find the value of \[\dfrac{dy}{dx}\]

Answer 1

Hint: We solve this problem first by applying the logarithm function on both sides to remove the power. We use the theorem that is
\[\log {{a}^{b}}=b\log a\]
Then we differentiate with respect to \['x'\] on both sides to get the value of \[\dfrac{dy}{dx}\]
For finding the value of \[\dfrac{dy}{dx}\] we use the product rule and chain rule.
The product rule states that if \[u,v\] are some functions then
\[\dfrac{d}{dx}\left( u\times v \right)=u\dfrac{dv}{dx}+v\dfrac{du}{dx}\]
The chain rule says that
\[\dfrac{d}{dx}\left( f\left( g\left( x \right) \right) \right)={f}'\left( g\left( x \right) \right).{g}'\left( x \right)\]

Complete step-by-step solution
We are given that the equation of \[x,y\] as
\[{{x}^{y}}={{y}^{x}}\]
Now, let us apply logarithm function on both sides then we get
\[\Rightarrow \log {{x}^{y}}=\log {{y}^{x}}\]
We know that the theorem of logarithm that is
\[\log {{a}^{b}}=b\log a\]
By using this theorem in above equation we get
\[\Rightarrow x\log y=y\log x\]
Now, by differentiating with respect to \['x'\] on both sides we get
\[\Rightarrow \dfrac{d}{dx}\left( x\log y \right)=\dfrac{d}{dx}\left( y.\log x \right)\]
We know that the product rule of differentiation that if \[u,v\] are some functions then
\[\dfrac{d}{dx}\left( u\times v \right)=u\dfrac{dv}{dx}+v\dfrac{du}{dx}\]
By using the product rule in above equation we get
\[\Rightarrow x\dfrac{d}{dx}\left( \log y \right)+\log y\dfrac{dx}{dx}=y.\dfrac{d}{dx}\left( \log x \right)+\log x\dfrac{dy}{dx}.......equation(i)\]
We know that the standard formula of differentiation that is
\[\Rightarrow \dfrac{d}{dx}\left( \log x \right)=\dfrac{1}{x}\]
We also know that the chain rule of differentiation that is
\[\dfrac{d}{dx}\left( f\left( g\left( x \right) \right) \right)={f}'\left( g\left( x \right) \right).{g}'\left( x \right)\]
By using the chain rule and the standard formula of differentiation in equation (i) we get
\[\Rightarrow x.\dfrac{1}{y}\dfrac{dy}{dx}+\log y=y.\dfrac{1}{x}+\log x.\dfrac{dy}{dx}\]
Now, let us take the \[\dfrac{dy}{dx}\] terms one side and remaining terms to other side then we get
\[\begin{align}
  & \Rightarrow \dfrac{dy}{dx}\left( \dfrac{x}{y}-\log x \right)=\dfrac{y}{x}-\log y \\
 & \Rightarrow \dfrac{dy}{dx}\left( \dfrac{x-y\log x}{y} \right)=\dfrac{y-x\log y}{x} \\
\end{align}\]
Now, by cross multiplying the terms we get
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{y}{x}\left( \dfrac{y-x\log y}{x-y\log x} \right)\]
Therefore the value of \[\dfrac{dy}{dx}\] is given as
\[\therefore \dfrac{dy}{dx}=\dfrac{y}{x}\left( \dfrac{y-x\log y}{x-y\log x} \right)\]

Note: We can solve this problem in another method.
We are given that the equation of \[x,y\] as
\[{{x}^{y}}={{y}^{x}}\]
Now, by differentiating with respect to \['x'\] on both sides we get
\[\Rightarrow \dfrac{d}{dx}\left( {{x}^{y}} \right)=\dfrac{d}{dx}\left[ {{y}^{x}} \right]\]
If \[u,v\] are the two functions then the general formula of differentiation is given as
\[\dfrac{d}{dx}\left( {{u}^{v}} \right)=\dfrac{d}{dx}\left( {{u}^{v}}\text{ assuming }u\text{ as constant} \right)+\dfrac{d}{dx}\left( {{u}^{v}}\text{ assuming }v\text{ as constant} \right)\]
By using the above formula we get
\[\Rightarrow \dfrac{d}{dx}\left( {{x}^{y}}\text{ as }x\text{ as constant} \right)+\dfrac{d}{dx}\left( {{x}^{y}}\text{ as }y\text{ as constant} \right)=\dfrac{d}{dx}\left( {{y}^{x}}\text{ as }y\text{ as constant} \right)+\dfrac{d}{dx}\left( {{y}^{x}}\text{as }x\text{ as constant} \right)\]
Now we know that the general formula of differentiation that is
\[\dfrac{d}{dx}\left( {{a}^{x}} \right)={{a}^{x}}\log a\] Where \['a'\] is constant.
By using this formula and chain rule to above equation we get
\[\begin{align}
  & \Rightarrow {{x}^{y}}\log x\dfrac{dy}{dx}+y.{{x}^{y-1}}={{y}^{x}}\log y+x.{{y}^{x-1}}.\dfrac{dy}{dx} \\
 & \Rightarrow {{x}^{y}}\left( \log x.\dfrac{dy}{dx}+\dfrac{y}{x} \right)={{y}^{x}}\left( \log y+\dfrac{x}{y}\dfrac{dy}{dx} \right) \\
\end{align}\]
We know that \[{{x}^{y}}={{y}^{x}}\] by substituting this in the above equation we get
\[\Rightarrow {{x}^{y}}\left( \log x.\dfrac{dy}{dx}+\dfrac{y}{x} \right)={{x}^{y}}\left( \log y+\dfrac{x}{y}\dfrac{dy}{dx} \right)\]
Now, let us take the \[\dfrac{dy}{dx}\] terms one side and remaining terms to other side then we get
\[\begin{align}
  & \Rightarrow \dfrac{dy}{dx}\left( \dfrac{x}{y}-\log x \right)=\dfrac{y}{x}-\log y \\
 & \Rightarrow \dfrac{dy}{dx}\left( \dfrac{x-y\log x}{y} \right)=\dfrac{y-x\log y}{x} \\
\end{align}\]
Now, by cross multiplying the terms we get
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{y}{x}\left( \dfrac{y-x\log y}{x-y\log x} \right)\]
Therefore the value of \[\dfrac{dy}{dx}\] is given as
\[\therefore \dfrac{dy}{dx}=\dfrac{y}{x}\left( \dfrac{y-x\log y}{x-y\log x} \right)\]