# Implicit Differentiation While an *explicit curve* is one defined using a $x \to y$ function, we can say that an *implicit curve* is one formed by a set of all points that satisfy a certain property. A basic example is that of a circle, which is defined by $x^2 + y^2 = r^2$. Here we don't have $y$ isolated on one side, instead the curve is defined as a set of all points that satisfy the above property. Although determining a derivative of a function defined implicitly using traditional methods is certainly possible (we'd solve for $y$ and determine the derivative explicitly), that method is often cumbersome. Instead, implicit differentiation can be used instead. In implicit differentiation we introduce two implicit functions that correspond to the left and right hand side of the equality, and we apply [[The Derivative Chain Rule]] to help identify the desired derivative. When doing implicit differentiation it is important to keep track of the implicit function's input and outputs. Careful when applying the chain rule! ## Example When determining the derivative of $3y^2 + 2x^3 = 9x$, we can define functions $j(x) = 3y^2 + 2x^3$ and $k(x) = 9x$. $y$ is the result of $f(x)$. Because the two functions are equal, their derivatives are equal as well, so we can write (as per [[The Derivative Chain Rule]]): $ \begin{align} \frac d{dx}[3 \cdot f(x)^2 + 2x^3] &= \frac d {dx}[9x] \\ 6 \cdot f(x)\cdot f'(x) + 6x^2 &= 9 \\ f'(x) &= \frac {9-6x^2}{6 \cdot f(x)} \\ f'(x) &= \frac {3-2x^2}{2y} \end{align} $ If we use the Leibniz's derivative notation, this becomes even simpler: $ \begin{align} 3y^2+2x^3 &=9x \\ 6y \frac {dy}{dx} + 6x^2 &= 9 \\ \frac {dy}{dx} &= \frac {9-6x^2} {6y} \\ \frac {dy}{dx}&= \frac {3-2x^2}{2y} \end{align} $