Example : $x^2+y^2=5$

To express the implicit function as an explicit function of $x$, we need to solve the equation for $y$. Then, we get $y=\pm \sqrt{25-x^2}$. The graph of two functions $y=\sqrt{25-x^2}$ and $y=-\sqrt{25-x^2}$ are the upper and lower semicircles of the circle $x^2+y^2=25$ as shown in Figures below.

Differentiation

In the case of the differentiation of implicit functions, we differentiate both sides of the equation with respect to x by using the Chain Rule, and solve the resulting equation for $y^{\prime }$.

Example: Find $\frac{dy}{dx}$ if $x^2+y^2=25$.

Solution : Differentiate both side of the equation:

$$\frac{d}{dx}(x^2+y^2)=\frac{d}{dx}(25)$$

Since $y$ is a function of $x$, applying the Chain Rule, we get

$$\frac{d}{dx}\left(y^2\right)=\frac{d}{dy}\left(y^2\right)\frac{dy}{dx}=2y\frac{dy}{dx}$$

Thus we have

$$2x+2y\frac{dy}{dx}=0$$

Solving the equation for $\frac{dy}{dx}$, we obtain

$$y^{\prime }=\frac{dy}{dx}=-\frac{x}{y}$$

NOTE: The expression $\frac{dy}{dx}=-\frac{x}{y}$ is the derivative in terms of both $x$ and $y$. If we want to express in terms of $x$, we only consider the functions $y$ given explicitly $y=\pm \sqrt{25-x^2}$.

More precisely,

• for $y=\sqrt{25-x^2}$, we have $\frac{dy}{dx}=-\frac{x}{y}=-\frac{x}{\sqrt{25-x^2}}$
• for $y=-\sqrt{25-x^2}$, we have $\frac{dy}{dx}=-\frac{x}{y}=\frac{x}{\sqrt{25-x^2}}$