Implicit and inverse-function differentiation Knowledge-Based Study

Study notes

1. Understanding Explicit vs. Implicit Relations

   An equation is written in explicit form when one variable is solved entirely in terms of another, such as $y = f(x)$. However, many mathematical curves are expressed as relations where $y$ cannot be easily isolated, such as the unit circle $x^2 + y^2 = 1$ or the Folium of Descartes $x^3 + y^3 = 6xy$. Such relationships define $y$ implicitly as an unknown function of $x$.

   The Chain Rule Mechanism

   Implicit differentiation uses the Chain Rule to differentiate both sides of an equation with respect to the independent variable $x$. When differentiating an expression containing $y$, we treat $y$ as an implicit function $y(x)$. Consequently, applying the Chain Rule gives:

   $$\frac{d}{dx}[y^n] = n y^{n-1} \cdot \frac{dy}{dx}$$

   Similarly, for composite expressions involving $y$, such as $\sin(y)$ or $e^y$, the derivatives are $\cos(y) \frac{dy}{dx}$ and $e^y \frac{dy}{dx}$ respectively.

   Systematic Procedure

   1. Differentiate both sides of the relation with respect to $x$, applying the Product, Quotient, and Chain Rules where appropriate.
   2. Collect every term containing the factor $\frac{dy}{dx}$ on one side of the equation.
   3. Move all terms that do not contain $\frac{dy}{dx}$ to the opposite side.
   4. Factor out $\frac{dy}{dx}$ and divide to solve explicitly for $\frac{dy}{dx}$.

   Finding Slopes and Tangent Lines

   Once $\frac{dy}{dx}$ is obtained in terms of $x$ and $y$, the numerical slope at any point $(x_0, y_0)$ lying on the curve is determined by directly substituting both coordinates into the expression for $\frac{dy}{dx}$.

   For example, consider the ellipse defined by $x^2 + 4y^2 = 8$. Differentiating both sides with respect to $x$:

   $$2x + 8y \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{x}{4y}$$

   At the point $(2, 1)$, the slope of the tangent line is $m = -\frac{2}{4(1)} = -\frac{1}{2}$. Using the point-slope formula, the tangent line equation is $y - 1 = -\frac{1}{2}(x - 2)$, or $x + 2y = 4$.

   Second-Order Implicit Derivatives

   To compute $\frac{d^2y}{dx^2}$, differentiate the first derivative $\frac{dy}{dx}$ with respect to $x$. Whenever $\frac{dy}{dx}$ appears in the resulting expression, substitute the formula established for the first derivative, and simplify using the original curve equation when possible.

   Theoretical Foundation

   Let $f$ be a one-to-one function that is differentiable on an interval $I$, with non-zero derivative $f'(x) \neq 0$. Its inverse function $g = f^{-1}$ satisfies the identity:

   $$f(g(x)) = x$$

   Differentiating both sides with respect to $x$ using the Chain Rule yields:

   $$f'(g(x)) \cdot g'(x) = 1$$

   Solving for $g'(x)$ gives the fundamental Inverse Function Derivative Formula:

   $$(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}$$

   Geometrical Interpretation

   The graph of $y = f^{-1}(x)$ is the reflection of $y = f(x)$ across the line $y = x$. If the tangent line to $f$ at $(a, b)$ has slope $m$, then the tangent line to $f^{-1}$ at the reflected point $(b, a)$ has slope $\frac{1}{m}$, provided $m \neq 0$.

   Deriving Derivatives via Implicit Differentiation

   The inverse function theorem and implicit differentiation work in harmony to determine derivatives of inverse trigonometric functions. Consider $y = \arcsin(x)$, defined for $x \in [-1, 1]$ and $y \in [-\frac{\pi}{2}, \frac{\pi}{2}]$.

   Rewriting this relation implicitly gives $\sin(y) = x$. Differentiating both sides with respect to $x$:

   $$\cos(y) \frac{dy}{dx} = 1 \implies \frac{dy}{dx} = \frac{1}{\cos(y)}$$

   Using the Pythagorean identity $\cos^2(y) + \sin^2(y) = 1$, and noting that $\cos(y) \ge 0$ on $[-\frac{\pi}{2}, \frac{\pi}{2}]$:

   $$\cos(y) = \sqrt{1 - \sin^2(y)} = \sqrt{1 - x^2}$$

   Thus, we establish the standard derivative rule:

   $$\frac{d}{dx}[\arcsin(x)] = \frac{1}{\sqrt{1 - x^2}}, \quad |x| < 1$$

   Key Formulas

   • $\frac{d}{dx}[\arccos(x)] = -\frac{1}{\sqrt{1 - x^2}}$
   • $\frac{d}{dx}[\arctan(x)] = \frac{1}{1 + x^2}$
   • $\frac{d}{dx}[\operatorname{arcsec}(x)] = \frac{1}{|x|\sqrt{x^2 - 1}}$