How do you use implicit differentiation to find \[\dfrac{dy}{dx}\] given \[{{x}^{2}}+{{y}^{2}}=2\]?

Hint: Differentiate both sides of the given function with respect to the variable x. Use the formula: - \[\dfrac{d{{x}^{n}}}{dx}=n{{x}^{n-1}}\] to simplify the L.H.S. Use the chain rule of differentiation to find the derivative of \[{{y}^{2}}\]. In the R.H.S. use the factor that the derivative of a constant is 0 to simplify. Keep the expression \[\dfrac{dy}{dx}\] in the L.H.S. and send the other variables and expressions to the R.H.S. to get the answer.

Complete step-by-step solution:
Here, we have been provided with the relation: - \[{{x}^{2}}+{{y}^{2}}=2\] and we are asked to find the value of \[\dfrac{dy}{dx}\].
\[\because {{x}^{2}}+{{y}^{2}}=2\]
Now, differentiating both the sides of the above relation with respect to the variable x, we get,
\[\Rightarrow \dfrac{d\left( {{x}^{2}}+{{y}^{2}} \right)}{dx}=\dfrac{d\left( 2 \right)}{dx}\]
Breaking the terms in the L.H.S., we get,
\[\Rightarrow \dfrac{d{{x}^{2}}}{dx}+\dfrac{d{{y}^{2}}}{dx}=\dfrac{d\left( 2 \right)}{dx}\]
Now, using the formula \[\dfrac{d{{x}^{n}}}{dx}=n{{x}^{n-1}}\], we get,
\[\Rightarrow 2x+\dfrac{d{{y}^{2}}}{dx}=\dfrac{d\left( 2 \right)}{dx}\]
We know that the derivative of a constant term is 0, so in the R.H.S. we must have 0,
\[\Rightarrow 2x+\dfrac{d{{y}^{2}}}{dx}=0\]
Applying the chain rule of differentiation to find the derivative of \[\dfrac{d{{y}^{2}}}{dx}\], we have,
\[\Rightarrow 2x+\dfrac{d{{y}^{2}}}{dy}\times \dfrac{dy}{dx}=0\]
What we are doing is, first we are differentiating \[{{y}^{2}}\] with respect to y and then we are differentiating y with respect to x and their product is considered. So, we have,
\[\Rightarrow 2x+2y\dfrac{dy}{dx}=0\]
Dividing both the sides with 2, we get,
\[\begin{align}
  & \Rightarrow x+y\dfrac{dy}{dx}=0 \\
 & \Rightarrow y\dfrac{dy}{dx}=-x \\
\end{align}\]
Dividing both the sides with y, we get,
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{-x}{y}\]
Hence, the derivative of the given relation is \[\dfrac{-x}{y}\].

Note: One may note that we can further simplify the relation that we have obtained, i.e., \[\dfrac{dy}{dx}=\dfrac{-x}{y}\], by substituting the value of y in terms of x. It can be written as: - \[y=\sqrt{-{{x}^{2}}+2}\]. In this way we will get the value of \[\dfrac{dy}{dx}\] in terms of only one variable, i.e., x. However, it is not of much importance here. You must remember the different rules of differentiation like: - the product rule, chain rule, \[\dfrac{u}{v}\] rule etc. Because they are common rules and are used everywhere in calculus.