2.6: Implicit Differentiation

This section explores how to take the derivative of implicit equations.

Differentiating Implicitly

Thus far, we have been given functions in the form $y = f(x)$, such as \[ y = x^2 \qquad y = \dfrac{t + 1}{t-1} \qquad y = \sqrt{36 - x}\]

These functions are in explicit form, meaning the dependent variable is in the form $y = f(x)$, where $y$ is isolated and the other side is independent variables.

Not all functions are of the form $y = f(x)$. Consider \[x^2y + y - x^2 + 1 = 0\] In thi form, $y$ is not explicitly a function of $x$, meaning $y$ is not isolated on one side. We call this an implicit equation.

Sometimes it is possible to convert an implicit equation into an explicit equation. From above we see that \begin{align} x^2y + y - x^2 + 1 &= 0 \\ y(x^2 + 1) &= x^2 - 1 \\ y &= \dfrac{x^2 - 1}{x^2 + 1} \end{align} and we have converted the implicit equation into an explicit equation.

This is not always possible. Consider \[y^4 - y^3 - y + 2x^3 - x = 8\] There is no way to isolate a single $y$ on one side.

In this case, we want to assume we can take $y = f(x)$ under suitable conditions. These suitable conditions usually mean to force the implicit equation to pass the vertical line test by chopping it up into multiple functions:

Finding the derivative of an implicit equation is called implicit differentiation.

Given $y^2 = x$, find $\dfrac{dy}{dx}$.

In general:

Finding $\dfrac{dy}{dx}$ with Implicit Differentiation
  1. Apply $\dfrac{d}{dx}$ to both sides.
  2. Use the chain rule on terms involving $y$.
  3. Differentiate terms involving $x$ normally.
  4. Solve the equation for $\dfrac{dy}{dx}$.
Given \[y^3 - y + 2x^3 - x = 8\] find $\dfrac{dy}{dx}$.
Given $x^2 + y^2 = 4$:
  1. Find $\dfrac{dy}{dx}$ using implicit differentiation.
  2. Find the slope of the tangent line at $(1, \sqrt{3})$ in the above graph.

Sometimes we need to use the product rule on terms involving both $x$ and $y$, such as $x^2y^3$.

Find $\dfrac{dy}{dx}$ for the following equation \[x^2y^3 + 6x^2 = \sin(y) + 12\]

Remember, if you want to isolate a variable, let's say "$x$", there are four steps:

  1. Expand all expressions into terms so there are no parentheses.
  2. Collect all terms with $x$ on one side. Put all other terms on the other.
  3. Convert $x$ into a factor by using the GCF factoring method.
  4. Divide both sides by the factor attached to $x$, therefore isolating $x$.

The next two problems involves product and chain rules.

Find the tangent line to the equation \[x^3 + y^3 = 6xy\] at $(3,3)$.
Find $dy/dx$ for the equation \[\sqrt{xy} = x + y\]
First, convert the square root into an exponent. Applying $\dfrac{d}{dx}$ gives \begin{align} \dfrac{d}{dx}\left[(xy)^{\frac{1}{2}}\right] &= \dfrac{d}{dx}\left[x + y\right] \\\dfrac{1}{2}(xy)^{-\frac{1}{2}}\cdot \dfrac{d}{dx}[xy] &= \dfrac{d}{dx}[x] + \dfrac{d}{dx}[y] \\\dfrac{1}{2}(xy)^{-\frac{1}{2}}(x\dfrac{d}{dx}[y] + y \dfrac{d}{dx}[x]) &= 1 + 1\cdot \dfrac{dy}{dx} \\\dfrac{1}{2}(xy)^{-\frac{1}{2}}(x\cdot 1\cdot \dfrac{dy}{dx} + y\cdot 1) &= 1 + \dfrac{dy}{dx} \\\dfrac{1}{2}(xy)^{-\frac{1}{2}}x\dfrac{dy}{dx} + \dfrac{1}{2}(xy)^{-\frac{1}{2}}y &= 1 + \dfrac{dy}{dx} \\\dfrac{1}{2}(xy)^{-\frac{1}{2}}x\dfrac{dy}{dx} - \dfrac{dy}{dx} &= 1 - \dfrac{1}{2}(xy)^{-\frac{1}{2}}y \\\dfrac{dy}{dx}\left(\dfrac{1}{2}(xy)^{-\frac{1}{2}}x - 1\right) &= 1 - \dfrac{1}{2}(xy)^{-\frac{1}{2}}y \\\dfrac{dy}{dx} &= \dfrac{1 - \dfrac{1}{2}\frac{1}{(xy)^{1/2}}y}{\dfrac{1}{2}\frac{1}{(xy)^{1/2}}x - 1} \\\dfrac{dy}{dx} &= \dfrac{1 - \dfrac{y}{2\sqrt{xy}}}{\dfrac{x}{2\sqrt{xy}} - 1}\cdot{\color{yellow} \dfrac{2\sqrt{xy}}{2\sqrt{xy}}} \\\dfrac{dy}{dx} &= \dfrac{2\sqrt{xy} - y}{x - 2\sqrt{xy}} \end{align}