Question

Consider the given expression, ${{x}^{y}}+{{y}^{x}}=2$ , then find the value of $\dfrac{dy}{dx}$.

Answer 1

Hint:We have to differentiate the equation with respect to x taking ${{x}^{y}}$ as u and ${{y}^{x}}$as v. Then we need to use product rule of differentiation to find the values .

Complete step-by-step answer:
We are given an equation which states that
${{x}^{y}}+{{y}^{x}}=2$
In order to find $\dfrac{dy}{dx}$ of the equation,
Let us consider ${{x}^{y}}$ = u and ${{y}^{x}}$ = v.
Solving ${{x}^{y}}$ = u …… (i)
Let us apply log on both sides of the equation.
log u = y log x
Differentiating both sides with respect to x, we get
$\dfrac{d}{dx}(log\text{ }u)=\dfrac{d}{dx}(y\log x)$
Now we know, differentiation of $\dfrac{d(\log y)}{dx}=\dfrac{1}{y}\dfrac{dy}{dx}$ , so above expression can be written as,
$\dfrac{1}{u}\dfrac{du}{dx}=\dfrac{d}{dx}(y\log x)$
Applying product rule, i.e., $\dfrac{d}{dx}\left( u.v \right)=u\dfrac{d}{dx}v+v\dfrac{d}{dx}u$, the above expression can be written as

& \dfrac{1}{u}\dfrac{du}{dx}=y\dfrac{d}{dx}(\log x)+\log x\dfrac{d}{dx}(y) \\\ & \Rightarrow \dfrac{1}{u}\dfrac{du}{dx}=\dfrac{dy}{dx}\log x+y.\dfrac{1}{x} \\\ \end{aligned}$$ Taking ‘u’ to the right hand side, we get $\dfrac{du}{dx}=u\left( \dfrac{dy}{dx}\log x+\dfrac{y}{x} \right)$ Substituting value from equation (i), we have $\dfrac{du}{dx}={{x}^{y}}\left( \dfrac{dy}{dx}\log x+\dfrac{y}{x} \right)$ …..(ii) Now consider, v = ${{y}^{x}}$ …..(iii) Taking log on both sides, we get log v = x log y Differentiating with respect to x, we get $$\dfrac{d}{dx}(log\text{ v})=\dfrac{d}{dx}(x\log y)$$ Now we know, differentiation of $\dfrac{d(\log y)}{dx}=\dfrac{1}{y}\dfrac{dy}{dx}$ , so above expression can be written as, $\dfrac{1}{v}\dfrac{dv}{dx}=\dfrac{d}{dx}\left( x\log y \right)$ Applying product rule, i.e., $$\dfrac{d}{dx}\left( u.v \right)=u\dfrac{d}{dx}v+v\dfrac{d}{dx}u$$, the above expression can be written as $$\begin{aligned} & \dfrac{1}{v}\dfrac{dv}{dx}=x\dfrac{d}{dx}(\log y)+\log y\dfrac{d}{dx}(x) \\\ & \Rightarrow \dfrac{1}{v}\dfrac{dv}{dx}=\log y+\dfrac{x}{y}\dfrac{dy}{dx} \\\ \end{aligned}$$ Taking ‘v’ on other side of equation, we get $\dfrac{dv}{dx}=v\left( \log y+\dfrac{x}{y}\dfrac{dy}{dx} \right)$ Substituting value from equation (iii) we get, $\dfrac{dv}{dx}={{y}^{x}}\left( \log y+\dfrac{x}{y}\dfrac{dy}{dx} \right)$ ….. (iv) From the question we know, ${{x}^{y}}+{{y}^{x}}=2$ u + v = 2 (As u = ${{x}^{y}}$ and v =${{y}^{x}}$) Differentiating the equation with respect to x, we get $\dfrac{du}{dx}+\dfrac{dv}{dx}=\dfrac{d(2)}{dx}$ From equation (ii) and (iv), we get ${{y}^{x}}\left( \log y+\dfrac{x}{y}\dfrac{dy}{dx} \right)+{{x}^{y}}\left( \dfrac{dy}{dx}\log x+\dfrac{y}{x} \right)=0$ (Differentiation of constant is zero) Opening the bracket, we get ${{y}^{x}}\log y+x.\dfrac{{{y}^{x}}}{y}\dfrac{dy}{dx}+{{x}^{y}}\dfrac{dy}{dx}\log x+y.\dfrac{{{x}^{y}}}{x}=0$ Grouping the terms, we get $\dfrac{dy}{dx}\left( x.\dfrac{{{y}^{x}}}{y}+{{x}^{y}}\log x \right)=-y.\dfrac{{{x}^{y}}}{x}-{{y}^{x}}\log y$ Taking the coefficient of $\dfrac{dy}{dx}$ on other side of equation, we get $\dfrac{dy}{dx}$ = -$\dfrac{y\dfrac{{{x}^{y}}}{x}+{{y}^{x}}\log y}{x\dfrac{{{y}^{x}}}{y}+{{x}^{y}}\log x}$ We know $\dfrac{{{x}^{a}}}{x}={{x}^{a-1}}$, applying this in the equation, we get $\dfrac{dy}{dx}=-\dfrac{y{{x}^{y-1}}+{{y}^{x}}\log y}{x{{y}^{x-1}}+{{x}^{y}}\log x}$ The required value of $\dfrac{dy}{dx}=-\dfrac{y{{x}^{y-1}}+{{y}^{x}}\log y}{x{{y}^{x-1}}+{{x}^{y}}\log x}$ Note: There are few steps in this equation where students usually commit a mistake. 1\. Here y is not a constant , so we cannot use the formula of ${{x}^{a}}$ and ${{a}^{x}}$ where differentiation of ${{x}^{a}}$ as ${{x}^{a-1}}$ and differentiation of ${{a}^{x}}$ as ${{a}^{x}}\ln a$. 2\. We must not forget to use the product rule of differentiation when two or more terms are present.