Question

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

Answer 1

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

& \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}$$ Now, by cross multiplying the terms we get $$\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{aligned} & \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{aligned}$$ We know that $${{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{aligned} & \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}$$ Now, by cross multiplying the terms we get $$\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)$$