Question

How do you find $\dfrac{dy}{dx}$ given $2xy+{{y}^{2}}=x+y$?

Answer 1

Hint: To find the value of $\dfrac{dy}{dx}$ we need to differentiate the given equation with respect to x. we will use the power rule and product rule to solve the differentiation. The product rule and power rule of differentiation are given as:
$\Rightarrow \dfrac{d}{dx}\left( xy \right)=x\dfrac{d}{dx}y+y\dfrac{d}{dx}x$
$\Rightarrow \dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}$

Complete step by step solution:
We have been given an equation $2xy+{{y}^{2}}=x+y$.
We have to find the value of $\dfrac{dy}{dx}$.
Now, we know that we have a function $y=f(x)$ where x is an independent variable and y is a dependent variable. If both dependent and independent variables are present in the equation then the function is implicit.
To solve the given equation we have to differentiate the both sides of the equation with respect to x. then we will get
$\Rightarrow \dfrac{d}{dx}\left( 2xy \right)+\dfrac{d}{dx}{{y}^{2}}=\dfrac{d}{dx}x+\dfrac{d}{dx}y$
Now, we know that $\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}$ and $\dfrac{d}{dx}\left( xy \right)=x\dfrac{d}{dx}y+y\dfrac{d}{dx}x$
Now, by applying the above formulas to the obtained equation we will get
\[\begin{align}
  & \Rightarrow 2\dfrac{d}{dx}\left( xy \right)+2{{y}^{2-1}}\dfrac{dy}{dx}=1+\dfrac{dy}{dx} \\
 & \Rightarrow 2\left( x\dfrac{d}{dx}y+y\dfrac{d}{dx}x \right)+2y\dfrac{dy}{dx}=1+\dfrac{dy}{dx} \\
\end{align}\]
Now, simplifying the above obtained equation we will get
\[\begin{align}
  & \Rightarrow 2\left( x\dfrac{dy}{dx}+y \right)+2y\dfrac{dy}{dx}=1+\dfrac{dy}{dx} \\
 & \Rightarrow 2x\dfrac{dy}{dx}+2y+2y\dfrac{dy}{dx}=1+\dfrac{dy}{dx} \\
\end{align}\]
Now, taking common terms out we will get
$\begin{align}
  & \Rightarrow 2x\dfrac{dy}{dx}+2y\dfrac{dy}{dx}-\dfrac{dy}{dx}=1-2y \\
 & \Rightarrow \dfrac{dy}{dx}\left( 2x+2y-1 \right)=1-2y \\
\end{align}$
Now, rearrange the terms to get the value of \[\dfrac{dy}{dx}\] we will get
$\Rightarrow \dfrac{dy}{dx}=\dfrac{1-2y}{2x+2y-1}$
Hence above is the required value of \[\dfrac{dy}{dx}\].

Note:
The point to be noted is that here in this question the given function is implicit function. So we have to differentiate the equation using chain rule. The point to be remembered is that in implicit differentiation we have to differentiate each term. The differentiation of y with respect to x in implicit differentiation is \[\dfrac{dy}{dx}\].