Question

How do you differentiate ${{x}^{2}}+{{y}^{2}}=2xy$ ?

Answer 1

We will answer this question by applying differentiation on both sides of the given equation using the basic concepts of differentiation formulae $\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}$ and $\dfrac{d}{dx}\left( uv \right)=u\dfrac{d}{dx}v+v\dfrac{d}{dx}u$ where $u$ and $v$ are functions of $x$ . Assume $y$ as constant and then simplify it and come to a conclusion.

Complete step by step answer:
For answering this question we need to differentiate the given expression ${{x}^{2}}+{{y}^{2}}=2xy$.
From the basic concepts of differentiation we have the formulae $\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}$ and $\dfrac{d}{dx}\left( uv \right)=u\dfrac{d}{dx}v+v\dfrac{d}{dx}u$ where $u$ and $v$ are the functions of $x$ .
After using these formulae and differentiate the given expression ${{x}^{2}}+{{y}^{2}}=2xy$ we will have $\dfrac{d}{dx}\left( {{x}^{2}}+{{y}^{2}}=2xy \right)$ .
By simplifying this equations we will have
$\begin{aligned} & \dfrac{d}{dx}\left( {{x}^{2}}+{{y}^{2}}=2xy \right) \\\ & \Rightarrow 2x+\dfrac{d}{dx}\left( {{y}^{2}} \right)=2y+2x\left( \dfrac{d}{dx}y \right) \\\ \end{aligned}$
By further simplifying we will have $2x+2y\dfrac{d}{dx}y=2y+2x\left( \dfrac{d}{dx}y \right)$ .
If $y$ is constant then $\dfrac{d}{dx}y$ will be zero otherwise we cannot predict its value without knowing the value of $y$ so we will keep it as it is.
Let us assume that $y$ is constant then we will have $\dfrac{d}{dx}y=0$ after using this in the above expression we will have $2x=2y\Rightarrow x=y$ .
Hence we can conclude that the differentiation of ${{x}^{2}}+{{y}^{2}}=2xy$ is $2x+2y\dfrac{d}{dx}y=2y+2x\left( \dfrac{d}{dx}y \right)$ if $y$ is constant we can say that $x=y$ by further simplifying the given equation.

Note: We should be careful while performing calculations for answering this question. If incase we had made a mistake and write it as $2x+2\dfrac{d}{dx}y=2y+2x\left( \dfrac{d}{dx}y \right)$ then we had made a mistake in this question this mistake would not make any difference but if $y$ is a function of $x$ then it will surely make a difference.