Theorem 3.8.1.
Let \(f(x) = x^{p/q}\) Then, \(f'(x) = \frac{p}{q}x^{p/q-1}\text{.}\)
Section 3.8 Implicit Differentiation
Subsection 3.8.1 Curves Given by Implicitly-Defined Functions
The slope of the curve \(y = f(x)\) is given by the derivative \(f'(x)\text{.}\) However, not all curves are defined by explicit functions, where \(y\) is isolated and is given as a function of \(x\text{.}\) Some functions are implicit, defined by a relation between \(x\) and \(y\text{.}\) More generally, curves in the plane are graphs of an equation in two variables, i.e. an equation of the form,
\begin{equation*}
F(x,y) = 0
\end{equation*}
where \(F(x,y)\) is a function of \(x\) and \(y\text{.}\)
For example, the circle with radius 1 centered at the origin has equation,
\begin{equation*}
x^2 + y^2 = 1
\end{equation*}
In some cases, it is possible to solve for \(y\) explicitly as an explicit function (or possibly multiple) of \(x\text{.}\) However, in other cases it is difficult or impossible to do so. In this case, it is not necessary to solve explicitly for \(y\) in order to determine the derivative of \(y\) with respect to \(x\text{.}\) Instead, we can use a technique called implicit differentiation.
Subsection 3.8.2 Implicit Differentiation
The idea of implicit differentiation is to differentiate both sides of an equation with respect to \(x\text{,}\) regarding \(y\) as a function of \(x\) with derivative \(\frac{dy}{dx} = y'\text{.}\) In other words, apply the differentiation operator \(\frac{d}{dx}\) to both sides of the equation. Then, solve the equation for \(\frac{dy}{dx}\) to determine an equation for the derivative of the function.
When differentiating an expression in \(y\text{,}\) it is necessary to use the chain rule.
When using implicit differentiation, \(\frac{dy}{dx}\) is usually in terms of both \(x\) and \(y\text{,}\) and so it is necessary to know both coordinates to determine the derivative at a point.
Subsection 3.8.3 Circles
Consider the unit circle \(x^2 + y^2 = 1\text{.}\)
Subsection 3.8.4 Horizontal Parabolas
Find \(\frac{dy}{dx}\) if \(y^2 = x\text{.}\) This is a horizontal parabola. It defines two functions of \(x\text{,}\) \(y_1(x) = \sqrt{x}\) and \(y_2(x) = -\sqrt{x}\text{.}\) Then, for \(x \neq 0\text{,}\)
\begin{equation*}
y_1'(x) = \frac{1}{2\sqrt{x}} \qquad \text{and} \qquad y_2'(x) = -\frac{1}{2\sqrt{x}}
\end{equation*}
Instead, differentiate both sides of the equation with respect to \(x\text{,}\)
\begin{align*}
\frac{d}{dx}(y^2) \amp = \frac{d}{dx}(x)\\
2y \frac{dy}{dx} \amp = 1\\
\frac{dy}{dx} \amp = \frac{1}{2y} \amp\amp \text{solving for }
\end{align*}
Notice that this covers derivatives for both the explicit solutions.
Subsection 3.8.5 Folium of Descartes
The equation \(x^3 + y^3 - 3xy = 0\) represents a curve called the folium of Descartes. Since \(y\) is a cubic polynomial in terms of \(x\text{,}\) it is possible to solve for \(y\) explicitly using the cubic formula, resulting in 3 explicit functions. However, the explicit expression is quite complicated, making the explicit derivative tedious to compute.
Subsection 3.8.6 Ellipses
Subsection 3.8.7 Hyperbolas
Consider a hyperbola, with equation of the form,
\begin{equation*}
\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1
\end{equation*}
This is a horizontal hyperbola. Differentiating implicitly,
\begin{equation*}
\frac{2x}{a^2} + \frac{2y y'}{b^2} \amp = 0
\end{equation*}
Solving for \(y'\text{,}\)
\begin{equation*}
y' = \frac{b^2 x}{a^2 y}
\end{equation*}
Then, the tangent line at a point \((x_0,y_0)\) on the parabola is given by,
\begin{equation*}
y - y_0 = \frac{b^2 x_0}{a^2 y_0} (x - x_0)
\end{equation*}
It turns out that this can be simplified further. This comes from trying to rewrite the equation into the standard form of a line \(Ax + By = C\text{.}\) We have,
\begin{align*}
(y - y_0) a^2 y_0 \amp = b^2 x_0 (x - x_0)\\
a^2 y_0 y - a^2 y_0^2 \amp = b^2 x_0 x - b^2 x_0^2\\
b^2 x_0 x - a^2 y_0 y \amp = b^2 x_0^2 - a^2 y_0^2\\
\frac{x_0 x}{a^2} - \frac{y_0 y}{b^2} \amp = \frac{x_0^2}{a^2} - \frac{y_0^2}{b^2}
\end{align*}
Then, since $(x_0, y_0)$ is a point on the hyperbola, it satisfies the equation $\frac{x_0^2}{a^2} - \frac{y_0^2}{b^2} = 1$. Thus, the equation of the tangent line simplifies to,
\begin{equation*}
\frac{x_0 x}{a^2} - \frac{y_0 y}{b^2} = 1
\end{equation*}
Subsection 3.8.8 Implicit Second Derivatives
After determining \(\frac{dy}{dx}\text{,}\) we can differentiate again to determine \(\frac{d^2y}{dx^2}\text{.}\) Typically, after simplifying, we can simplify by substituting the equation of the curve.
Subsection 3.8.9 Remark on Where Implicit Differentiation Fails
Implicit differentiation cannot always be applied. When using implicit differentiation, in particular when differentiating an equation involving \(y\text{,}\) it is implicitly assumed (see what I did there) that \(y\) is indeed a differentiable function with respect to \(x\text{.}\) However, this need not be the case. For example, the circle of radius \(\sqrt{R}\text{,}\)
\begin{equation*}
x^2 + y^2 = R
\end{equation*}
Using implicit differentiation, like above we get,
\begin{equation*}
\frac{dy}{dx} = -\frac{x}{y}
\end{equation*}
This determines the slope at any point on the curve, except when \(y = 0\text{.}\) However, if \(R = 0\text{,}\) then \(x^2 + y^2 = 0\) only represents a single point, the origin \((0,0)\text{,}\) for which the concept of slope is meaningless. If \(R \lt 0\text{,}\) then \(x^2 + y^2 = R\) represents no points in the plane (much less a curve), so \(\frac{dy}{dx}\) is meaningless here also. In general, being able to calculate \(\frac{dy}{dx}\) using implicit differentiation does not guarantee that it represents the slope of anything.
Subsection 3.8.10 Implicit Function Theorem
The implicit function theorem gives sufficient conditions for implicit differentiation to find the slope of the graph of an implicit function.
Subsection 3.8.11 Power Rule for Rational Exponents Proof
Implicit differentiation, along with the power rule for integers, can be used to prove the power rule for rational exponents.
Proof.
If \(y = x^{p/q}\text{,}\) then \(y^q = x^p\text{.}\) Then, using implicit differentiation,
\begin{equation*}
\frac{d}{dx} y^q \amp = \frac{d}{dx} x^p
\end{equation*}
By the power rule (for integer powers),
\begin{equation*}
qy^{q-1} \frac{dy}{dx} \amp = px^{p-1}
\end{equation*}
Then, solving for \(\frac{dy}{dx}\text{,}\)
\begin{equation*}
\frac{dy}{dx} = \frac{px^{p-1}}{qy^{q-1}}
\end{equation*}
Finally, substituting back \(y = x^{p/q}\text{,}\)
\begin{align*}
\frac{dy}{dx} \amp = \frac{px^{p-1}}{q(x^{p/q})^{q-1}}\\
\amp = \frac{p}{q} \frac{x^{p-1}}{x^{p-p/q}}\\
\amp = \frac{p}{q} x^{p/q-1} \amp\amp \text{laws of exponents}
\end{align*}
