Implicit Differentiation

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We discuss how to take a derivative of an implicitly defined function.

Video Lecture

There is a supplemental video on understanding when and why implicit differentiation can be used! You can find the video here:

https://youtu.be/yvIXTkwhXJ4

(Supplemental Videos are included via external link so you don’t have to watch them to earn credit.)

Text and Additional Details

Classically we are used to functions being defined in a way that makes the output obvious. For example; $y = 3x^2 - 12$ has a clear input variable, $x$ , and a clear output variable $y$ , and most importantly it is very easy to get the value of $y$ when you plug in $x$ . In this case we would say the function is “explicitly defined” in terms of $x$ .

But not all functions are defined like this. Indeed, as you may have seen in precalculus, you can have implicitly defined functions, which is when the dependent variable (aka output) is no longer isolated like it was in the previous example. For instance, consider the implicitly defined function $y^2 = 3xy$ . Here, $y$ is still the output variable, and $x$ is still the input variable, but after substituting a value for $x$ it takes more effort to determine what the value of $y$ would be.

As we move forward in calculus then, we want to know how to handle these kinds of functions as well.

It’s important to realize that implicitly defined functions are still essentially the same kind of object as explicitly defined functions in many ways; indeed the one (rather notable) difference is that our dependent variable isn’t isolated. This difference can take a lot of forms. One possibility is that our dependent variable is only on one side of the equation, but with operations being performed on it - for example $y^2 = 3x + 1$ . Another possibility is that the dependent and independent variable are intermixed, like in $y = x^2 + 2xy + y^2$ .

This is where our normal notation kind of hurts us though. We write the dependent variable as if it is just a variable... but in reality it is a function of an independent variable (this is what distinguishes the implicitly defined expression $3xy + y^2$ from the explicitly defined multivariable expression $3xy + y^2$ ... two different things that nonetheless look totally identical - very unfortunate). For this reason, it can be useful to write $y(x)$ instead of just $y$ to make it clear that $y$ is really a function being applied to an input $x$ .

Why does this matter for us? Because when we see $y(x)$ this should look very similar (and indeed is functionally the same as) $f(x)$ ; and we know how to deal with taking a derivative of $f(x)$ ... it’s just $f'(x)$ .

This seemingly simple observation is the heart of implicit differentiation. When we have implicitly defined functions, we can take the derivative as normal, but we need to keep track of where the derivative is being applied to the dependent variable. Let’s look at the example $y^2 = 3xy$ .

We are going to do this problem in two parallel columns to see what it looks like to do the “implicit differentiation” part, versus how we might think of doing it if we replaced “ $y$ ” with the “ $f(x)$ ” style notation. We will use $u$ in the “Notes” section as a stand-in for both $y$ and $f(x)$ notations (so if you see $u$ , you can sub in either $y$ or $f(x)$ to see what we did in the corresponding column for that line).


$\displaystyle y$ -Version	$\displaystyle f(x)$ -Version	Notes

$\displaystyle y^2 = 3xy$	$\displaystyle (f(x))^2 = 3x\cdot f(x)$
$\displaystyle \frac {d}{dx} \left [ y^2 \right ] = \frac {d}{dx} \left [ 3xy \right ]$	$\displaystyle \frac {d}{dx} \left [ \left (f(x)\right )^2 \right ] = \frac {d}{dx} \left [ 3xf(x) \right ]$	Apply Derivative
$\displaystyle 2(y) \cdot \left (\frac {d}{dx} \left [ y \right ]\right ) = \frac {d}{dx} \left [ 3xy \right ]$	$\displaystyle 2f(x)\cdot \frac {d}{dx} \left [ f(x) \right ] = \frac {d}{dx} \left [ 3xf(x) \right ]$	Derivative of $u^2=2u$
		and chain rule.
$\displaystyle 2(y) \cdot y' = \frac {d}{dx} \left [ 3xy \right ]$	$\displaystyle 2f(x)\cdot f'(x) = \frac {d}{dx} \left [ 3xf(x) \right ]$	$\frac {d}{dx} u(x) = u'(x)$
$\displaystyle 2(y) \cdot y' = \frac {d}{dx} \left [ 3x \right ]\cdot y + (3x) \cdot \frac {d}{dx} \left [ y \right ]$	$\displaystyle 2f(x)\cdot f'(x) = \frac {d}{dx} \left [ 3x \right ]\cdot f(x) + (3x) \cdot \frac {d}{dx} \left [ f(x) \right ]$	Apply Product Rule
$\displaystyle 2(y) \cdot y' = 3 \cdot y + (3x) \cdot \frac {d}{dx} \left [ y \right ]$	$\displaystyle 2f(x)\cdot f'(x) = 3 \cdot f(x) + (3x) \cdot \frac {d}{dx} \left [ f(x) \right ]$	$\frac {d}{dx} [3u] = 3$
$\displaystyle 2(y) \cdot y' = 3y + 3x \cdot y'$	$\displaystyle 2f(x)\cdot f'(x) = 3 f(x) + 3x \cdot f'(x)$	$\frac {d}{dx} u(x) = u'(x)$

As we can see, the idea of implicit differentiation is nothing but the chain rule, applied to the fact that the dependent variable $y$ is really a disguised version of the more familiar function notation $f(x)$ - so when we write something like $y'$ we really mean $f'(x)$ .

It is important to note that it is expected of students in this segment to solve for the derivative term. For example, if we continue from where we left off with the $y$ version above, we want to isolate $y'$ ...

$\displaystyle \frac {d}{dx} \left [ y^2 \right ] = \frac {d}{dx} \left [ 3xy \right ]$
$\displaystyle 2(y) \cdot y' = 3y + 3x \cdot y'$	From previous work.
$\displaystyle 2(y) \cdot y' - 3x \cdot y' = 3y$	Subtract $\displaystyle 3x \cdot y'$ from both sides.
$\displaystyle y' \left ( 2y - 3x \right ) = 3y$	Factor out $y'$ from both terms on the left.
$\displaystyle y' = \frac {3y}{2y - 3x}$	Divide both sides by $2y - 3x$ .

So we can see we have isolated the derivative term (the $y'$ ). Notice that it still has $y$ mixed into the derivative - this is pretty normal. Since we started with an implicitly defined function, it is natural that the derivative would also be implicitly defined.

When you need to take a derivative of an implicitly defined function you need to...

Look up what to do. Take a derivative as normal. Take a derivative of the function, using the chain rule on the dependent variables which generate a “prime” version of those variables. Burn it with fire. That always works.

Once you have taken the derivative...

You’re done, on to the next problem! You need to solve for the derivative term, the “prime” version of the dependent variable. You need to compute the derivative of the prime terms to finish the derivative. More fire. Kill it with fire.

Finally, you may wonder how to know if a term is implicitly defined or not; meaning, when do you include something like a $y'$ or not? The good news is, the answer is entirely contained in the notation!

When we read something like $\frac {d}{dx}[x + y]$ , we usually think of this as “the derivative of $x+y$ ”, but this is missing an absolutely vital part of the notation. The derivative operator, the “ $\frac {d}{dx}$ ” symbol part is indeed telling us we want to take a derivative... but it is also telling us what we are taking the derivative with respect to. In particular, the $\color {red}x$ in $\frac {d}{d{\color {red}x}}$ is telling you that $x$ is the variable that gets treated as the independent variable, and any other variable will be treated as a dependent variable. Thus to compute this we would have the following:

$\frac {d}{dx}[x + y]$	$=\frac {d}{dx}[x] + \frac {d}{dx}[y]$	Derivative Sum Rule
	$= 1 + \frac {d}{dx}[y]$	Since the Derivative was with respect to $x$ , the derivative of $x$ is $1$ .
	$= 1 + 1 \cdot y'$	Since the Derivative was with respect to $x$ , we first treat $y$ as if it were an “ $x$ ” to get $1$ ,
		but then we have to apply the chain rule and multiply by the derivative of $y$ ; which is just $y'$ .
	$= 1 + y'$	Simplified the expression.

So, if the variable in the bottom of the $\frac {d}{du}$ (the “ $u$ ”) matches the variable you are taking a derivative of, you don’t have any chain rule “prime” terms; but if they don’t match, you should end up with the chain-rule giving you a “prime” version of the variable multiplying the expression. Here’s a quick few examples:

$\displaystyle \frac {d}{da} a = 1$	$\displaystyle \frac {d}{da} b = b'$	$\displaystyle \frac {d}{da} c = c'$	$\displaystyle \frac {d}{da} x = x'$
$\displaystyle \frac {d}{db} a = a'$	$\displaystyle \frac {d}{db} b = 1$	$\displaystyle \frac {d}{db} c = c'$	$\displaystyle \frac {d}{db} x = x'$
$\displaystyle \frac {d}{dc} a = a'$	$\displaystyle \frac {d}{dc} b = b'$	$\displaystyle \frac {d}{dc} c = 1$	$\displaystyle \frac {d}{dc} x = x'$
$\displaystyle \frac {d}{dx} a = a'$	$\displaystyle \frac {d}{dx} b = b'$	$\displaystyle \frac {d}{dx} c = c'$	$\displaystyle \frac {d}{dx} x = 1$

The last column may look strange with the $x'$ since we are use to taking derivatives with respect to $x$ , but this just means that the derivative being taken is with respect to a variable other than $x$ and so the derivative assumes that $x$ is now the dependent variable and not an independent variable.

How do you know which letters need prime notation and which ones are just treated like normal?

If the letter matches the bottom of the derivative operator (the “u” variable in $\frac {d}{du}$ ) then treat it as normal, otherwise it gets the chain rule and a prime. If the letter matches the bottom of the derivative operator (the “u” variable in $\frac {d}{du}$ ) then give it a prime notation with chain rule, otherwise treat it as normal. There isn’t any way to tell unless you know which letters are dependent variables and which are independent variables. Uh... napalm? Why didn’t fire work?

As we’ve seen, implicit differentiation is nothing more than using the chain rule and recognizing that we sometimes use a dependent variable (like $y$ ) in lieu of the functional notation $f(x)$ . Using a variable like $y$ instead of $f(x)$ can make a lot of sense, and if you are interested in where or why you might run into implicitly defined functions in the real world, there is an optional video covering it (linked at the top of the page after the lecture video). However, the real takeaway here is to not let the notation trip you up - you want to keep track of what the derivative is respect to (say, $x$ ) and what the function is in terms of (say, $x$ , and $y$ ). Wherever these two don’t match up, you need to include a leftover prime notation due to the chain rule. So if you are taking $\frac {d}{dx}[y]$ , since the derivative is in terms of $x$ and the term you are differentiating is $y$ ... which is not an $x$ , you have a leftover $y'$ .