Calculus - Implicit Differentiation and Higher Derivatives

Implicit vs Explicit Functions: An explicit function is written in the form $y = f(x)$, where $y$ is isolated. An implicit function defines a relationship between $x$ and $y$ where $y$ is not isolated, such as the equation of a circle $x^2 + y^2 = r^2$. Visually, explicit functions always pass the vertical line test, while implicit relations like ellipses or foliums of Descartes may loop back on themselves, requiring implicit differentiation to find gradients at specific points.

The Chain Rule in Implicit Differentiation: When differentiating terms involving $y$ with respect to $x$, we must treat $y$ as a differentiable function of $x$. According to the chain rule, $\frac{d}{dx}[g(y)] = g'(y) \cdot \frac{dy}{dx}$. For example, the derivative of $y^2$ is not simply $2y$, but $2y\frac{dy}{dx}$.

Derivatives of Mixed Terms: Terms involving both $x$ and $y$ (like $xy$ or $x^2y^3$) require the application of the product rule or quotient rule alongside implicit differentiation. For example, $\frac{d}{dx}(xy) = x\frac{dy}{dx} + y(1)$. This creates an algebraic equation where you must group all $\frac{dy}{dx}$ terms on one side to solve for the derivative.

Higher-Order Derivatives: The second derivative, denoted as $\frac{d^2y}{dx^2}$ or $f''(x)$, represents the rate of change of the first derivative. It describes the 'acceleration' of a function or the curvature of a graph. If $y$ is the position, $\frac{dy}{dx}$ is velocity, and $\frac{d^2y}{dx^2}$ is acceleration.

Concavity and Geometry: The second derivative provides information about the shape of a curve. If $\frac{d^2y}{dx^2} > 0$ on an interval, the graph is concave up (it bends upwards like a cup). If $\frac{d^2y}{dx^2} < 0$, the graph is concave down (it bends downwards like a frown). A point where the concavity changes (and where $\frac{d^2y}{dx^2} = 0$ or is undefined) is called a point of inflection.

Finding Higher Derivatives Implicitly: To find the second derivative of an implicit function, differentiate the expression for $\frac{dy}{dx}$ with respect to $x$. This usually results in an expression containing $x$, $y$, and $\frac{dy}{dx}$. To find a final answer in terms of $x$ and $y$ only, you must substitute the original expression for $\frac{dy}{dx}$ back into the second derivative equation.