Implicit Differentiation

Section 1.1 Implicit Differentiation

Subsection 1.1.1 Implicit Curves and Their Slopes

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 $y\text{.}$ For example,

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\begin{align*} \begin{array}{c|c} \text{explicit} \amp \text{implicit} \\ \hline y = 2x + 5 \amp 2x + 3y = 6 \\ y = \sin{x} + 4 \amp x^2 + y^2 = 4 \end{array} \end{align*}

Basically,

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Explicit means $y$ is isolated.
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Implicit means the $y$ is not isolated, so the $x$ and $y$ terms are mixed together.
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Example 1.1.1.

For example, a circle with radius 1 centered at the origin (the unit circle) has the equation,

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\begin{gather*} x^2 + y^2 = 1 \end{gather*}

For the entire graph, this is not a function, because it fails the vertical line test. However, it is still something which we could create tangent lines for.

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We want to take derivatives of these kinds of curves. To do so, one strategy would be to isolate $y\text{,}$ and then take the derivative like we have done previously. In this case,

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\begin{align*} x^2 + y^2 \amp = 1\\ y^2 \amp = 1 - x^2 \amp\amp \text{subtracting $x^2$ from both sides}\\ y \amp = \pm \sqrt{1-x^2} \amp\amp \text{taking the square root of both sides} \end{align*}

This leads to two different functions, say $y_1(x) = \sqrt{1-x^2}$ and $y_2(x) = - \sqrt{1-x^2}\text{.}$ The first is for the top semicircle, and the second is for the bottom semicircle (Desmos link).

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For some equations, it is possible to solve for $y$ as an explicit function (or possibly multiple functions) of $x\text{.}$ However, in many other more complicated cases, this is difficult or actually algebraically impossible.

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In fact, it is not necessary to solve for $y$ explicitly to find the derivative. Instead, we can use a technique called implicit differentiation.

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Example 1.1.2. Motivational Example: Circle.

Consider the unit circle, with equation $x^2 + y^2 = 1\text{.}$ The idea is that we can take the derivative of both sides of the equation,

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\begin{align*} \frac{d}{dx} (x^2 + y^2) \amp = \frac{d}{dx} (1)\\ \frac{d}{dx}(x^2) + \frac{d}{dx} (y^2) \amp = \frac{d}{dx}(1) \end{align*}

As usual,

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$\frac{d}{dx} (x^2) = 2x$ (power rule) and,
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$\frac{d}{dx}(1) = 0$ (derivative of a constant is zero)
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However, $\frac{d}{dx}(y^2)$ is a bit different, because $y$ represents a function of $x\text{,}$ so instead of $y\text{,}$ we could more precisely write $y(x)\text{,}$ like,

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\begin{gather*} \frac{d}{dx} (y(x))^2 \end{gather*}

This means that $y^2$ is like a composition of function, with $y$ as the inner function, and the square is the outer function. This means we have to use the chain rule,

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\begin{gather*} \frac{d}{dx} (y(x))^2 = \underbrace{2 \cdot y(x)}_{\text{power rule}} \cdot \underbrace{\frac{dy}{dx}}_{\text{chain rule}} \end{gather*}

Then, the equation becomes,

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\begin{gather*} 2x + 2y \frac{dy}{dx} = 0 \end{gather*}

Then, we want to solve for the derivative $\frac{dy}{dx}\text{,}$ so we can isolate it in the equation,

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\begin{align*} 2y \frac{dy}{dx} \amp = -2x\\ \frac{dy}{dx} \amp = \frac{-2x}{2y}\\ \frac{dy}{dx} \amp = -\frac{x}{y} \end{align*}

This means that for any point $(x,y)$ on the unit circle, the slope of the tangent line at that point is $-\frac{x}{y}\text{.}$ For example, for the point $(x,y) = (\frac{3}{5}, \frac{4}{5})\text{,}$ the derivative (slope) at that point is,

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\begin{gather*} \frac{dy}{dx} = -\frac{3/5}{4/5} = -\frac{3}{4} \end{gather*}

Note that $y$ is included in the formula for the derivative (not just $x$), and this is totally normal and ok.

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Also, we can use the derivative formula to find where the graph has a horizontal tangent, and also where it has a vertical tangent.

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There is a horizontal tangent if $y' = 0\text{,}$ or $-\frac{x}{y} = 0\text{.}$ This means that $x = 0\text{.}$ Plugging this into the curve equation $x^2 + y^2 = 1\text{,}$ we get $y^2 = 1\text{,}$ so $y = \pm 1\text{.}$ So, $(0,1)$ and $(0,-1)\text{.}$ This should make sense graphically.
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For vertical tangents, this is basically where the derivative is undefined, because of division by 0. Here, it is when $y = 0\text{.}$ Plugging this into the equation, we get $x^2 = 1\text{,}$ so $x = \pm 1\text{,}$ and so the points are $(1,0)$ and $(-1,0)\text{.}$
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Example 1.1.3. Horizontal Parabolas.

Consider the equation $y^2 = x$ (or $x = y^2$). This equation represents a horizontal parabola. We could solve this equation for $y\text{,}$ by taking the square root of both sides to get $y = \pm \sqrt{x}\text{.}$ In other words, this defines two functions of $x\text{,}$ $y_1(x) = \sqrt{x}$ and $y_2(x) = -\sqrt{x}\text{.}$ We could take the derivative of each individually, to get,

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\begin{gather*} y_1'(x) = \frac{1}{2\sqrt{x}} \qquad \text{and} \qquad y_2'(x) = -\frac{1}{2\sqrt{x}} \end{gather*}

Instead, using implicit differentiation, we can take the derivative of both together. Differentiate both sides of the equation,

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\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} \quad \text{solving for $\frac{dy}{dx}$} \end{align*}

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Example 1.1.4. Folium of Descartes.

The equation $x^3 + y^3 = 3xy$ represents a curve called the folium of Descartes (“folium” is Latin for “leaf”). Notice that $x$ and $y$ are mixed together, so this is an implicit equation. It is technically possible to solve for $y$ in this formula, however this requires advanced techniques, and the explicit expression is very complicated, making the derivative tedious to compute. Instead, using implicit differentiation, take the derivative of both sides,

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\begin{align*} \frac{d}{dx} \brac{x^3 + y^3} \amp = \frac{d}{dx}(3xy)\\ 3x^2 + 3y^2 \cdot y' \amp = 3xy' + 3y \end{align*}

Then, solving for $y'\text{,}$

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\begin{align*} 3y^2 y' - 3xy' \amp = 3y - 3x^2\\ (3y^2 - 3x) y' \amp = 3y - 3x^2\\ y' \amp = \frac{3y - 3x^2}{3y^2 - 3x}\\ y' \amp = \frac{y - x^2}{y^2 - x} \end{align*}

Then, for example, for the point $(1,2)$ on the curve (you can verify that $(1,2)$ is indeed on the curve, by plugging it into the equation $x^3 + y^3 = 3xy\text{,}$ and verifying that both sides are equal),

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\begin{gather*} y' = \frac{2 - 1^2}{2^2 - 1} = \frac{1}{3} \end{gather*}

Then, the equation of the tangent line at $(1,2)$ is,

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\begin{gather*} y - 2 = \frac{1}{3}(x-1) \end{gather*}

We can also consider where on this curve has a horizontal tangent line. This occurs when $y' = 0\text{,}$ or,

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\begin{align*} \frac{y - x^2}{y^2 - x} \amp = 0 \end{align*}

This occurs when the numerator is equal to 0, or $y - x^2 = 0\text{,}$ or $y = x^2\text{.}$ This doesn’t give a particular point, but instead a relationship between $x$ and $y\text{.}$ Combining this with the original curve equation, this forms like a system of equations,

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\begin{align*} \begin{cases} y - x^2 = 0 \amp \text{for $y' = 0$} \\ x^3 + y^3 = 3xy \amp \text{for the point to be on the curve} \end{cases} \end{align*}

Then, solving this will give us the points where the tangent line is horizontal.

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\begin{align*} x^3 + (x^2)^3 \amp = 3x (x^2)\\ x^3 + x^6 \amp = 3x^3\\ x^6 - 2x^3 \amp = 0\\ x^3(x^3 - 2) \amp = 0\\ x = 0 \qquad \amp x = \sqrt[3]{2} \end{align*}

Then, substituting back into the equation $y = x^2\text{,}$ we get $y = 0$ and $y = \sqrt[3]{4}\text{.}$ Therefore, there is a horizontal tangent at $(0,0)$ and $(\sqrt[3]{2}, \sqrt[3]{4})\text{.}$

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Subsection 1.1.2 Summary of Implicit Differentiation

In summary, to do implicit differentiation:

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Differentiate both sides (with respect to $x$), i.e. “take $\frac{d}{dx}$ of both sides”.
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When differentiating any term involving $y\text{,}$ treat $y$ as a function of $x\text{,}$ and use the chain rule (basically, multiply by $y'$ at the end). Keep in mind that,
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\begin{gather*} \frac{d}{dx} x = 1 \qquad \text{and} \qquad \frac{d}{dx} y = y' \end{gather*}

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Solve for $y'$ (isolate $y'$) in the equation.
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Some notes:

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You can use either $y'$ or $\frac{dy}{dx}$ to represent the derivative of $y$ with respect to $x\text{.}$ The latter notation $y'$ is a bit more compact.
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With implicit differentiation, the derivative $\frac{dy}{dx}$ is usually in terms of both $x$ and $y\text{.}$ This means, to evaluate the derivative at a particular point, it is necessary to know both coordinates. In contrast, the derivative of an explicit function only requires $x\text{.}$
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We can think of the derivative $\frac{d}{dx}$ as an operator that we can apply to both sides of the equation. This is analogous to how we can square both sides, or take the logarithm of both sides.
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Implicit differentiation also has the advantage that it leads to one formula for the derivative, that applies to all points on the graph, even if the graph may have 2 or 3 or even more separate function equations.
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Subsection 1.1.3 Horizontal and Vertical Tangents

Horizontal tangents occur when $y' = 0\text{.}$
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Vertical tangents occur when $y'$ is undefined.
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For a curve with $y' = \frac{\text{numerator}}{\text{denominator}}\text{,}$

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Horizontal tangents when $\text{numerator} = 0\text{.}$
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Vertical tangents when $\text{denominator} = 0\text{.}$
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Note: technically, if both the numerator and denominator are 0, then this is a separate case and the regular rules don’t apply.

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