Question

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

Answer 1

Hint: Differentiate both sides of the given function with respect to the variable x. Use the formula: - $${\displaystyle \frac{d{x}^n}{dx}}=n{x}^{n-1}$$ 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 $${\displaystyle \frac{dy}{dx}}$$ 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 $${\displaystyle \frac{dy}{dx}}$$.
$$\because x^2+y^2=2$$
Now, differentiating both the sides of the above relation with respect to the variable x, we get,
$$\Rightarrow {\displaystyle \frac{d(x^2+y^2)}{dx}}={\displaystyle \frac{d\left(2\right)}{dx}}$$
Breaking the terms in the L.H.S., we get,
$$\Rightarrow {\displaystyle \frac{d{x}^2}{dx}}+{\displaystyle \frac{d{y}^2}{dx}}={\displaystyle \frac{d\left(2\right)}{dx}}$$
Now, using the formula $${\displaystyle \frac{d{x}^n}{dx}}=n{x}^{n-1}$$, we get,
$$\Rightarrow 2x+{\displaystyle \frac{d{y}^2}{dx}}={\displaystyle \frac{d\left(2\right)}{dx}}$$
We know that the derivative of a constant term is 0, so in the R.H.S. we must have 0,
$$\Rightarrow 2x+{\displaystyle \frac{d{y}^2}{dx}}=0$$
Applying the chain rule of differentiation to find the derivative of $${\displaystyle \frac{d{y}^2}{dx}}$$, we have,
$$\Rightarrow 2x+{\displaystyle \frac{d{y}^2}{dy}}\times {\displaystyle \frac{dy}{dx}}=0$$
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,
$$\Rightarrow 2x+2y{\displaystyle \frac{dy}{dx}}=0$$
Dividing both the sides with 2, we get,
$$\begin{array}{rl}& \Rightarrow x+y{\displaystyle \frac{dy}{dx}}=0\\ & \Rightarrow y{\displaystyle \frac{dy}{dx}}=-x\end{array}$$
Dividing both the sides with y, we get,
$$\Rightarrow {\displaystyle \frac{dy}{dx}}={\displaystyle \frac{-x}{y}}$$
Hence, the derivative of the given relation is $${\displaystyle \frac{-x}{y}}$$.

Note: One may note that we can further simplify the relation that we have obtained, i.e., $${\displaystyle \frac{dy}{dx}}={\displaystyle \frac{-x}{y}}$$, 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 $${\displaystyle \frac{dy}{dx}}$$ 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, $${\displaystyle \frac{u}{v}}$$ rule etc. Because they are common rules and are used everywhere in calculus.