Question

Differentiate the following:
\[{{\text{y}}^{x}}={{x}^{y}}\]

Answer 1

Hint: First of all, we should differentiate \[{{\text{y}}^{x}}={{x}^{y}}\] on both sides. Now we will apply the formula\[\log {{a}^{b}}=b\log a\]. Now we will differentiate on both sides. By using \[d(uv)=udv+vdu\], we can find the differentiation of both L.H.S and R.H.S. Now we should apply \[\dfrac{d}{dx}(\log x)=\dfrac{1}{x}\]for further steps. After this by taking \[\dfrac{dy}{dx}\] on one side and remaining terms on the other side. This will give us the value of \[\dfrac{dy}{dx}\] for the equation \[{{\text{y}}^{x}}={{x}^{y}}\].

Complete step by step solution:
Now we will differentiate the equation \[{{\text{y}}^{x}}={{x}^{y}}\] on both sides.
Now let us apply log on both sides.
\[\Rightarrow \log {{y}^{x}}=\log {{x}^{y}}\]
We know that \[\log {{a}^{b}}=b\log a\]. In the same way, we get
\[\Rightarrow x\log y=y\log x\]
Now, let us differentiate on both sides.
\[\Rightarrow \dfrac{d}{dx}(x\log y)=\dfrac{d}{dx}(y\log x)....(1)\]
We know that the \[d(uv)=udv+vdu\].
Now by using this rule we will solve the equation (1).
\[\Rightarrow x\dfrac{d}{dx}(\log y)+\log y\dfrac{dx}{dx}=y\dfrac{d}{dx}(\log x)+\log x\dfrac{dy}{dx}\]
We know that \[\dfrac{d}{dx}(\log x)=\dfrac{1}{x}\].
Now we will apply this formula.
\[\begin{align}
  & \Rightarrow x\left( \dfrac{1}{y} \right)\dfrac{dy}{dx}+\log y=y\left( \dfrac{1}{x} \right)+\operatorname{logx}\dfrac{dy}{dx} \\
 & \Rightarrow \dfrac{x}{y}\dfrac{dy}{dx}+\log y=\dfrac{y}{x}+\log x\dfrac{dy}{dx} \\
 & \Rightarrow \dfrac{x}{y}\dfrac{dy}{dx}-\log x\dfrac{dy}{dx}=\dfrac{y}{x}-\log y \\
 & \Rightarrow \dfrac{dy}{dx}\left( \dfrac{x}{y}-\log x \right)=\dfrac{y}{x}-\log y \\
 & \Rightarrow \dfrac{dy}{dx}\left( \dfrac{x-y\log x}{y} \right)=\dfrac{y-x\log y}{x} \\
\end{align}\]
By using cross multiplication, we get
\[\begin{align}
  & \Rightarrow \dfrac{dy}{dx}=\left( \dfrac{y-x\log y}{x} \right)\left( \dfrac{y}{x-y\log x} \right) \\
 & \Rightarrow \dfrac{dy}{dx}=\left( \dfrac{y}{x} \right)\left( \dfrac{y-x\log y}{x-y\operatorname{logx}} \right) \\
\end{align}\]
So, the differentiation of \[{{\text{y}}^{x}}={{x}^{y}}\] is equal to \[\left( \dfrac{y}{x} \right)\left( \dfrac{y-x\log y}{x-y\operatorname{logx}} \right)\].

Note: Students should use the formulae for differentiation carefully in this question. Students should also try to avoid calculation mistakes in this problem to get a correct answer.
Some students will have a misconception that
\[\dfrac{d({{y}^{x}})}{dx}={{y}^{x}}\log y.....(1)\]
Students may assume that while calculating the value of \[\dfrac{d({{y}^{x}})}{dx}\], they may consider y as constant.
\[\dfrac{d({{x}^{y}})}{dx}={{x}^{y-1}}.....(2)\]
Students may assume that while calculating the value of \[\dfrac{d({{x}^{y}})}{dx}\], they may consider x as constant. This is a totally incorrect process. In the problem, we were asked to find the differentiation of y with respect to x but equation (1) gives the partial differentiation of y with respect to x and equation (2) gives the partial differentiation of x with respect to y. So, students should have a clear view on the difference between differentiation and partial differentiation.