Question

How do you find the derivative of $x\ln y-y\ln x=1$?

Answer 1

This type of question is based on the concept of implicit differentiation. We can solve this question with the help of product rule of differentiation, that is,$\dfrac{d}{dx}\left( uv \right)=u\dfrac{dv}{dx}+v\dfrac{du}{dx}$ where u and v are the functions of ‘x’. Differentiate the given equation with the help of chain rule of differentiation $\dfrac{d}{dx}f\left( y \right)={f}'\left( y \right)\dfrac{dy}{dx}$. Since the differentiation of a constant is always 0, $\dfrac{d}{dx}\left( 1 \right)=0$. Equate the differentiation of $x\ln y-y\ln x$ with zero and then find the value of $\dfrac{dy}{dx}$.

Complete step-by-step answer:
According to the question, we have been given the equation as $x\ln y-y\ln x=1$.
We need to find the differentiation of $x\ln y-y\ln x=1$.

First, we have to consider the equation $x\ln y-y\ln x=1$ -----(1)
Differentiate equation (1) with respect to x
Therefore,
$\dfrac{d}{dx}\left( x\ln y-y\ln x \right)=\dfrac{d}{dx}\left( 1 \right)$
We know that differentiation of a constant is always 0.
We get,
$\dfrac{d}{dx}\left( x\ln y-y\ln x \right)=0$ -----(2)
Now using the addition rule of multiplication $\dfrac{d}{dx}\left( u+v \right)=\dfrac{du}{dx}+\dfrac{dv}{dx}$ in (2), we get,
$\dfrac{d}{dx}\left( x\ln y-y\ln x \right)=\dfrac{d}{dx}\left( x\ln y \right)+\dfrac{d}{dx}\left( -y\ln x \right)$
$\Rightarrow \dfrac{d}{dx}\left( x\ln y \right)-\dfrac{d}{dx}\left( y\ln x \right)=0$ -------(3)

Now let us consider $\dfrac{d}{dx}\left( x\ln y \right)$.
Using the product rule of differentiation $\dfrac{d}{dx}\left( uv \right)=u\dfrac{dv}{dx}+v\dfrac{du}{dx}$, we get,
$\dfrac{d}{dx}\left( x\ln y \right)=x\dfrac{d}{dx}\left( \ln y \right)+\ln y\dfrac{dx}{dx}$
Now, let us use chain rule of differentiation $\dfrac{d}{dx}f\left( y \right)={f}'\left( y \right)\dfrac{dy}{dx}$ in $\dfrac{d}{dx}\left( \ln y \right)$ and $\dfrac{d}{dx}\left( \ln x \right)=\dfrac{1}{x}$, we get,
$\dfrac{d}{dx}\left( x\ln y \right)=x\dfrac{1}{y}\dfrac{dy}{dx}+\ln y\dfrac{dx}{dx}$
$\Rightarrow \dfrac{d}{dx}\left( x\ln y \right)=x\dfrac{1}{y}\dfrac{dy}{dx}+\ln y$
$\Rightarrow \dfrac{d}{dx}\left( x\ln y \right)=\dfrac{x}{y}\dfrac{dy}{dx}+\ln y$

Now let us consider $\dfrac{d}{dx}\left( y\ln x \right)$.
Using the product rule of differentiation $\dfrac{d}{dx}\left( uv \right)=u\dfrac{dv}{dx}+v\dfrac{du}{dx}$, we get,
$\dfrac{d}{dx}\left( y\ln x \right)=y\dfrac{d}{dx}\left( \ln x \right)+\ln x\dfrac{dy}{dx}$
Now, let us use $\dfrac{d}{dx}\left( \ln x \right)=\dfrac{1}{x}$, we get,
$\dfrac{d}{dx}\left( y\ln x \right)=y\dfrac{1}{x}+\ln x\dfrac{dy}{dx}$
$\Rightarrow \dfrac{d}{dx}\left( y\ln x \right)=\dfrac{y}{x}+\ln x\dfrac{dy}{dx}$

Substituting these values in equation (3), we get,
$\dfrac{d}{dx}\left( x\ln y \right)-\dfrac{d}{dx}\left( y\ln x \right)=0$
$\Rightarrow \dfrac{x}{y}\dfrac{dy}{dx}+\ln y-\left( \dfrac{y}{x}+\ln x\dfrac{dy}{dx} \right)=0$
On further simplification, we get,
$\Rightarrow \dfrac{x}{y}\dfrac{dy}{dx}+\ln y-\dfrac{y}{x}-\ln x\dfrac{dy}{dx}=0$
Now, group all the $\dfrac{dy}{dx}$ to the left-hand side and the rest to the right-hand side.
$\Rightarrow \dfrac{x}{y}\dfrac{dy}{dx}-\ln x\dfrac{dy}{dx}=\dfrac{y}{x}-\ln y$
Taking $\dfrac{dy}{dx}$ common from the obtained equation, we get,
$\Rightarrow \dfrac{dy}{dx}\left( \dfrac{x}{y}-\ln x \right)=\dfrac{y}{x}-\ln y$
$\Rightarrow \dfrac{dy}{dx}=\dfrac{\dfrac{y}{x}-\ln y}{\dfrac{x}{y}-\ln x}$
On taking LCM, we get,
$\Rightarrow \dfrac{dy}{dx}=\dfrac{\dfrac{y-x\ln y}{x}}{\dfrac{x-y\ln x}{y}}$
$\therefore \dfrac{dy}{dx}=\dfrac{y\left( y-x\ln y \right)}{x\left( x-y\ln x \right)}$

Hence, the derivative of $x\ln y-y\ln x=1$ is $\dfrac{dy}{dx}=\dfrac{y\left( y-x\ln y \right)}{x\left( x-y\ln x \right)}$.

Note: Whenever we get this type of problems, we need to make sure about the rules and properties of differentiation. In this question, we use the differentiation of logarithmic functions. Also, we should avoid calculation mistakes to obtain accurate answers. Similarly, we should not get confused with the sign conventions. Chain rule and product rule of differentiation should be used, if needed.