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.

& \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{aligned}$$ By using cross multiplication, we get $$\begin{aligned} & \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{aligned}$$ So, the differentiation of $${{\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.