Question

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

Answer 1

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,

& \Rightarrow x+y\dfrac{dy}{dx}=0 \\\ & \Rightarrow y\dfrac{dy}{dx}=-x \\\ \end{aligned}$$ Dividing both the sides with y, we get, $$\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.