Implicit Differentiation

Lesson Features »

Lesson Priority: High

Calculus $\longrightarrow$
Discovering Derivatives $\longrightarrow$
  • Understand implicit differentiation as an extension of the chain rule
  • Understand how implicit differentiation gives us generic results for unknown functions of $x$
  • Recognize when we do and do not necessarily need to use this approach
Lesson Description

In some ways, regular differentiation is just a subset of implicit differentiation. Implicit can also be viewed as an extension of the chain rule. This lesson shows you what it's all about, but most importantly, gives you the know-how to get the right answer for exams and future applications.

Practice Problems

Practice problems and worksheet coming soon!

You Should Know
Implicit Differentiation becomes a thorn in the side of so many students, because it is both peculiar and particular relative to many other derivatives topics. Fortunately, it is a topic that is brutally predictable in terms of what your professor expects you to do on a test - and because of that, putting yourself through the ringer by drilling practice problems for 30 to 60 minutes should set you completely straight in terms of preparing to take an exam! This is not how things play out in most Calculus topics - furthermore is not a substitute for actual understanding, but for better or worse, it is a practical one.

Working with the Unknown

Implicit differentiation is often a thorn in a Calc students side, because it's both odd to understand and particular to execute. At its heart, it exists so that we might better quantify derivatives of functions that aren't known or can't be expressed in the classic $y=$ form of functions.Right off the bat that sounds a bit wonky - why would we care about taking the derivative of a function if we don't even know what it is? Leaving the philosophy aspect aside and looking at the how-to, it's worth noting that we've already done this - it's just another way of describing the Chain Rule ». Recall its definition:
Define: The Chain Rule (Revisited)The derivative of a composition of functions is calculated as follows:$$\frac{d}{dx} \,\, f\big(g(x)\big) = f'\big(g(x)\big) \cdot g'(x)$$
While it's true that we typically use and quote the Chain Rule in instances where we know the function we're working with, it's possible to use it in cases where we don't. Let's see an example.
Example 1Let $y$ be an unknown function of $x$. Using the Chain Rule, describe the derivative of $y^2$.$\blacktriangleright$ For ease of notation we are referring to this function as $y$, but keep in mind that it's really $y(x)$, since we said it's a function of $x$. Therefore, the derivative we're being asked to find is really$$\frac{d}{dx} \,\, \left[y(x)\right]^2$$which, by the chain rule, is$$\frac{d}{dx} \,\, [y(x)]^2 = 2[y(x)] \cdot y'(x)$$We have just computed the derivative of $y^2$ with respect to $x$ even though we don't know what $y$ is. If we did know what $y$ is, we could have turned $y^2$ into an $x$ expression, and moved forward with derivative processes that we're already familiar with.
Define: Chain Rule for Unknown y = y(x)Let $y = y(x)$ be an unknown function of $x$. By the Chain Rule, it follows that$$\frac{d}{dx} \, y = \frac{dy}{dx}$$Furthermore,$$\frac{d}{dx} \, g(y) = g'(y) \cdot \frac{dy}{dx}$$
Again, this definition is nothing more than an adaptation of the Chain Rule.Let's see one more example of a derivative involving an unknown function that requires the Chain Rule, and then verify its results with a known function's explicit derivative.
Example 2 (Unknown $y$)Find the derivative of $2y + y^2$ with respect to $x$, where $y$ is an unknown function of $x$.$\blacktriangleright$ Using the Chain Rule and derivative rules on each term, we know that:$$\frac{d}{dx} \, 2y = 2 \frac{d}{dx} \, y = 2 \frac{dy}{dx}$$$$\frac{d}{dx} \, y^2 = 2y \, \frac{dy}{dx}$$Therefore,$$\frac{d}{dx} \, \big[2y + y^2\big] = 2 \frac{dy}{dx} + 2y \, \frac{dy}{dx}$$
Example 2 (Explicit)Find the derivative of $2y + y^2$, for the function $y = 2e^{3x}$.$\blacktriangleright$ Since we already figured out that the implicit derivative of $2y + y^2$ is $2y' + 2y \, y'$, let's use that result, knowing that $y = 2e^{3x}$ and therefore $y' = 6e^{3x}$.$$\frac{d}{dx} \, \big[ 2y + y^2 \big] = 2y' + 2y \, y'$$$$ = 2\big(6e^{3x}\big) + 2\big(2e^{3x}\big)\big(6e^{3x}\big)$$$$ = 12e^{3x} + 24e^{6x}$$Now, to check our work, let's take the derivative of the expression explicitly. To do this, replace $y$ with $2e^{3x}$ before taking any derivatives.$$2y + y^2 = 2(2e^{3x}) + (2e^{3x})^2 = 4e^{3x} + 4e^{6x}$$$$\frac{d}{dx} \, \big[4e^{3x} + 4e^{6x}\big] = 12e^{3x} + 24e^{6x}$$It's a match! So why do we bother with implicit processes in the first place? As we're about to see, implicit differentiation allows us to find an expression for $y'$ even in relationships that defy traditional one-variable function definitions.

What Do We Want To Actually Do?

The goal of most implicit differentiation problems is simply this: get an expression for $y'$ (or often we'll use differential notation $dy/dx$). This is valuable in $x$-$y$ relationships that are not easily written in traditional $y=$ language, but rather are defined by an equation involving both $x$ and $y$. In short, we will be able to figure out an expression for $y'$ without specifying an explicit definition of the function in $y=$ form.The major steps for this process are:
  1. Take the implicit derivative of both sides of the equation by applying the derivative operator $\frac{d}{dx}$
  2. Take the derivative of each term with respect to $x$
  3. Move all the $dy/dx$ terms to one side of the equation, all other terms to the opposite side
  4. Isolate the $dy/dx$ term via algebra manipulation
Let's see an example.
Example 3Consider the "sideways" parabola $x = y^2 + 3y - 4$.Find the expression that describes the tangent slope on this graph at any given point $(x,y)$.$\blacktriangleright$ Since we measure slope in the coordinate plane as $\Delta y / \Delta x$, $\frac{dy}{dx}$ is the expression that will give us the instantaneous "rise over run" slope we want at any given point. We can find $dy/dx$ for this equation without isolating $y$ by using Implicit Differentiation.First, apply the derivative operator $d/dx$ to both sides, treating it like a multiplicative and distributing as necessary.$$\frac{d}{dx} \, \big[ x \big] = \frac{d}{dx} \, \big[ y^2 + 3y - 4 \big]$$$$\frac{d}{dx} \, x = \frac{d}{dx} \, y^2 + \frac{d}{dx} \, 3y - \frac{d}{dx} \, 4$$Consider each term in turn.$$\frac{d}{dx} \, x = 1$$because we're taking the $x$ derivative of an $x$ expression, so it is no different than derivatives we've been learning up to this point.$$\frac{d}{dx} \, y^2 = 2y \, \frac{dy}{dx}$$This derivative is based on the Chain Rule as discussed above, because $y$ depends on $x$.$$\frac{d}{dx} \, 3y = 3 \, \frac{dy}{dx}$$Similarly to the previous derivative, the Chain Rule applies.$$\frac{d}{dx} \, 4 = 0$$There is no $x$ or $y$ variable in this term, and the derivative of a constant on its own is always $0$.When we put all these pieces together, we have$$1 = 2y \, \frac{dy}{dx} + 3 \, \frac{dy}{dx}$$To finish the task, we will solve for $dy/dx$. Treat it like one big variable that you're solving for.$$1 = \frac{dy}{dx} \, \big[ 2y + 3 \big]$$$$\frac{1}{2y+3} = \frac{dy}{dx}$$We factored it out from each term on the right side, and then divided to isolate it. The result is the answer we seek. That is to say,$$\frac{dy}{dx} = \frac{1}{2y+3}$$is the expression that describes the tangent slope of this equation at any point $(x,y)$ on its graph.Let's look at another example.
Example 4Find $dy/dx$ for the relationship $y = 2x^2 + \sin(y)$, where $y$ is an unknown function of $x$.$\blacktriangleright$ Any attempt to write this relationship as a $y=$ style function with $y$ isolated on one side will be in vain because it's algebraically not possible to do so. Instead, let's take the implicit derivative of each side of the equation.$$\frac{d}{dx} \, y = \frac{d}{dx} \, \bigg[ 2x^2 + \sin(y) \bigg]$$$$\frac{dy}{dx} = 4x + \cos(y) \frac{dy}{dx}$$As we've seen, derivatives of terms that have the unknown function $y$ will involve the Chain Rule factor of $dy/dx$. However, the derivative with respect to $x$ of $2x^2$ is $4x$ - this is because we are talking about taking an $x$ derivative of an $x$ expression.Next, let's move the terms with $dy/dx$ to the left side, and all other terms to the right side.$$\frac{dy}{dx} - \cos(y) \frac{dy}{dx} = 4x$$Finally, isolate $dy/dx$ and solve for it. To do this, factor it out like it's a variable.$$\frac{dy}{dx} \bigg(1 - \cos(y) \bigg) = 4x$$$$\therefore \frac{dy}{dx} = \frac{4x}{1 - \cos(y)}$$
You Should Know
If some of this still seems confusing (when the $dy/dx$ appears versus not, applying a derivative operator to both sides, treating $dy/dx$ like a variable, having $y$ in the answer), try not to get too hung up on the concept. Most students do much better to understand the how, more than the why, in terms of getting through exam questions. As we continue to look at the common problems we would expect to encounter on a test, I'll say honestly (and you won't often hear me say this) that if you can figure out how these problems work mechanically and mimic what I'm doing to be able to correctly work problems, you'll be fine for an exam even if you have no idea what's going on behind the scenes.
Pro Tip
Nearly every exam question you'll encounter on implicit differentiation will ask you to find $dy/dx$. This typically means starting with an $x$-$y$ equation, taking the implicit derivative of both sides, and isolating the $dy/dx$ term by treating it like a variable. If you become masterful at this, even if you have little-to-no idea as to what the heck you are doing conceptually, you will probably ace an exam on this topic.

Specific Tangent Slopes

Most of the time when we are asked to work problems, the answer we're asked to turn in is$$\frac{dy}{dx} = \mathrm{some} \,\, x \,\, \mathrm{and} \,\, y \,\, \mathrm{expression}$$because they ask us to "find $dy/dx$ for the given equation". However, it's also somewhat common (and not many more steps) to be asked to then take the result and calculate the actual tangent slope of the curve at a specified coordinate. The only thing we have to do after solving for $dy/dx$ is plug in the $x$ and $y$ values that are specified.Let's look again at Example 3.
Example 3 (Revisited)Calculate the slope of the curve $x = y^2 + 3y - 4$ at the point $(-4,-3)$. Then express the tangent line to the curve at that point in the linear form $y = mx + b$.$\blacktriangleright$ This is the same curve we examined in Example 3 - it is a "sideways" parabola that we cannot express as a $y=$ style function. However, using Implicit Differentiation, we found that its derivative is$$\frac{dy}{dx} = \frac{1}{2y+3}$$Therefore, its slope at the point $x=-4$ and $y=-3$ can be calculated by plugging in those values into the $dy/dx$ expression as needed:$$\frac{dy}{dx}\Bigg|^{(-4, -3)} = \frac{1}{2(-3)+3}$$$$=-\frac{1}{3}$$The slope of $x = y^2 + 3y - 4$ at the point $(-4,-3)$ is $-1/3$.Finally, knowing the slope, we can figure out the slope-intercept form of the tangent line by plugging in $(-4,-3)$ and $m=-1/3$ into the generic $y=mx+b$ form.$$y = mx + b$$$$-3 = -\frac{-4}{3} + b$$$$b = -\frac{13}{3}$$$$\therefore y = -\frac{x}{3} - \frac{13}{3}$$is the tangent line to the curve $x = y^2 + 3y - 4$ at the point $(-4,-3)$, expressed in slope-intercept form as we were asked to do. Here's a graph of this curve and tangent line to help us see that our answer looks reasonable.Let's look at one more example of finding a specific slope of a curve in a situation where Implicit Differentiation makes it possible.
Example 5A circle centered at the origin with radius $5$ is defined by the equation$$x^2 + y^2 = 25$$(1)Find the tangent slope of this circle at the point $(-2, -\sqrt{21})$.$\blacktriangleright$ Hopefully we remember a bit about equations of circles » in the coordinate plane - at least enough to not only recognize the equation form, but also quickly verify that the point $\left(-2, \sqrt{21}\right)$ is indeed on the circle.Depending on what you've seen in your past Algebra and Pre-Calc courses, you may also know that you can't express a circle as a traditional function in $y=$ form. Graphically, we know this to be true instantly because a circle fails the vertical line test ». However, using Implicit Differentiation, we can take the derivative of both sides of this equation with respect to $x$, and isolate the $dy/dx$ term. Whatever we figure out that $dy/dx$ is equal to will be the tangent slope that we usually think about for traditional functions, since we can interpret $dy/dx$ as "the instantaneous change in $y$ as $x$ changes", which is analogous to the traditional "rise over run" slope definition.$$\frac{d}{dx} \, \big[ x^2 + y^2 \big] = \frac{d}{dx} \, 25$$The implicit derivative of $x^2$ with respect to $x$ is $2x$, just as we would expect in previous lessons where we were taking $d/dx$ derivatives of $x$ expressions.The implicit derivative of $y^2$ with respect to $x$ is$$2y \, \frac{dy}{dx}$$This is because of the Chain Rule factor we've been needing throughout this lesson - measuring how the expression $y^2$ changes as $x$ changes is a Chain Rule derivative because $y$ depends on $x$.Finally, the derivative of $25$ with respect to $x$ is $0$, for the same reasons that derivatives of lone constants are always $0$.Plugging these results into equation (1) yields$$2x + 2y \, \frac{dy}{dx} = 0$$Now we'll solve for $dy/dx$:$$2y \, \frac{dy}{dx} = -2x$$$$\frac{dy}{dx} = \frac{-2x}{2y} = -\frac{x}{y}$$By isolating $dy/dx$ using our implicit process, we have found the expression for the tangent slope of the curve we started with. And since we were asked to calculate the slope at a particular point, we need only plug in the given $x$ and $y$ values.$$\frac{dy}{dx} \Bigg|^{(-2, -\sqrt{21})} = -\frac{-2}{-\sqrt{21}} = -\frac{2}{\sqrt{21}} \approx -0.436$$From a graph, we can see that this is a reasonable answer even though we can't count grid boxes exactly to check the slope.
Only half the battle with implicit differentiation is understanding conceptually what is going on. Most of my students find they are able to get through exams fairly well with a little how-to instruction and practice, without worrying about processing the concepts and reasons why this is all a big extension of the Chain Rule. If you can do the work, you'll get the credit - so get down to business, practice the common problems that we are expected to be able to work, and you'll be set straight in no time.

Mr. Math Makes It Mean

Product Rule ProblemsOnce you're grounded on the fundamental types of problems, it's time to look at the arguably most common (and unfortunately, most easy to screw up) type of implicit problem - product rule expressions.Let's learn by doing with a common example.
Example 6Find the implicit derivative of $y$ with respect to $x$ for the relationship $xy=4$.$\blacktriangleright$ As before, we will apply the $d/dx$ operator to both sides.$$\frac{d}{dx} \, xy = \frac{d}{dx} \, 4$$Consider the derivative of $xy$ with respect to $x$. Since $y$ is an object that itself depends on $x$, we have to treat it like a function of $x$ and therefore, as we have seen, its derivative is $dy/dx$. But since both $x$ and $y$ have derivatives and these objects are being multiplied together, we truly have a Product Rule situation on our hands. The derivative of the product of two objects must be computed using the Product Rule.Product Rule Reminder:$$\frac{d}{dx} \, \big[ f(x) \cdot g(x) \big]$$$$ = \left[ \frac{d}{dx} \, f(x) \right] g(x) + f(x) \left[ \frac{d}{dx} \, g(x) \right]$$Similarly, it follows that$$\frac{d}{dx} \, \big[xy\big]$$$$ = \left[ \frac{d}{dx} \, x \right] y + x \left[ \frac{d}{dx} \, y \right]$$$$=1 \cdot y + x \, \frac{dy}{dx} = y + x \, \frac{dy}{dx}$$Now, let's resume the problem, keeping in mind that $d/dx$ of $4$ is $0$.$$\frac{d}{dx} \, xy = \frac{d}{dx} \, 4$$$$\longrightarrow \, y + x \, \frac{dy}{dx} = 0$$We must isolate $dy/dx$. When we do, we'll have completed the task.$$x \, \frac{dy}{dx} = -y$$$$\frac{dy}{dx} = - \frac{y}{x}$$
Expressions that require the product rule in implicit processes are the number one killer of good grades on this topic. There's no sugar coating the fact that this takes some practice and understanding - it's unique and unlike anything you've seen thus far. The best advice I have it to take a step back and really understand why the product rule is required here, given that $y$ is not simply an independent variable, but rather a variable that in turn depends on $x$. The reason we need the Product Rule here is the same reason we need the Product Rule to differentiate something like $x^2 \cdot f(x)$.
Make sure to practice more of these Product Rule problems, both in the Put It To The Test section below, and in the accompanying practice worksheet for this lesson, as well as anywhere else you can find. You can recognize them quickly because they have $x$ and $y$ expressions multiplied together (e.g. $x^2 y$, $y\sin(x)$, $xe^{y}$, etc.).

Higher Derivatives

Higher derivatives must be calculated iteratively when we work with implicit derivatives, due to the unknown nature of the variable $y$ (which depends somehow on $x$). The process is always the same, though, albeit potentially tedious.
Define: Higher Implicit DerivativesTo find the second derivative of $y$ with respect to $x$ (notated $\frac{d^2 y}{dx^2}$ ) you must first find $\frac{dy}{dx}$ using the typical implicit approach. Then, apply the $\frac{d}{dx}$ operator to both sides of that result, noting that$$\frac{d}{dx}\, \frac{dy}{dx} = \frac{d^2 y}{dx^2}$$Important: when / if any $\frac{dy}{dx}$ terms appear on the other side of the equation, replace them with the $\frac{dy}{dx}$ expression you solved for previously.
This will make a lot more sense and be easier to follow with an example.
Example 7Calculate $\frac{d^2y}{dx^2}$ for the hyperbola $x^2 - y^2 = 16$, where the variable $y$ depends on $x$.$\blacktriangleright$ First, let's apply the $d/dx$ operator to both sides and proceed to figure out what $dy/dx$ is.$$x^2 - y^2 = 16$$(1)$$\frac{d}{dx} \, \big[ x^2 - y^2 \big] = \frac{d}{dx} \, 16$$$$2x - 2y \, \frac{dy}{dx} = 0$$$$\frac{dy}{dx} = \frac{x}{y}$$(2)To find the second derivative, we now apply the $d/dx$ operator to this result, as we said in the definition above.$$\frac{d}{dx} \, \frac{dy}{dx} = \frac{d}{dx} \, \frac{x}{y}$$The right side will require the quotient rule.$$\frac{d^2y}{dx^2} = \frac{y \, \big[\frac{d}{dx} \, x\big] - x \, \big[\frac{d}{dx} \, y \big]}{y^2}$$$$=\frac{y - x \frac{dy}{dx}}{y^2}$$(3)Also as stated in the definition, at this point, we need to replace any $dy/dx$ we have in our work (result 3) with the definition of $dy/dx$ (result 2).$$\frac{d^2y}{dx^2} = \frac{y - x \big( \frac{x}{y}\big)}{y^2}$$Clean this up algebraically to obtain the final answer:$$\frac{d^2y}{dx^2} = \frac{\frac{y^2-x^2}{y}}{y^2}$$$$=\frac{y^2 - x^2}{y^3}$$The process of finding second derivatives this way is tedious, but aside from asking something like Wolfram Alpha » to do your dirty work for you, there is no workaround or shortcut. Fortunately, because this question type is tedious, you're probably only going to see 1 or 2 of them on an exam, if at all.Let's try one more of these.
Example 8Find $d^2y/dx^2$ for$$3-xy = x + y$$$\blacktriangleright$ Once again, let's start by finding the first derivative of $y$ with respect to $x$.$$3-xy = x + y$$(1)$$\frac{d}{dx} \big[ 3 - xy \big] = \frac{d}{dx} \big[ x + y \big]$$(2)Don't forget that $d/dx \, xy$ requires the product rule:$$\frac{d}{dx} \, xy = x \, \frac{dy}{dx} + 1 \cdot y$$With that in mind, result (2) becomes$$0 - \big(x \, \frac{dy}{dx} + y \big) = 1 + \frac{dy}{dx}$$$$\Rightarrow \, -x \, \frac{dy}{dx} -y = 1 + \frac{dy}{dx}$$(3)Isolate $dy/dx$:$$-y - 1 = \frac{dy}{dx} + x \, \frac{dy}{dx}$$$$-y - 1 = \frac{dy}{dx} \bigg( 1 + x \bigg)$$Cleaning up, we have$$\frac{dy}{dx} = - \left(\frac{y + 1}{x + 1}\right)$$(4)Next, to proceed with finding the second derivative, let's take $d/dx$ of both sides of result (4), which will require the quotient rule on the right side.$$\frac{d}{dx} \, \frac{dy}{dx} = \frac{d}{dx} \, \left[-\frac{y + 1}{x + 1}\right]$$$$\frac{d^2 y}{dx^2} = -\frac{(x+1) \, \frac{dy}{dx} - (y+1)}{(x+1)^2}$$As we've said, when finding higher derivatives with implicit differentiation, we need to plug in for $dy/dx$ as we go, since we have already computed it (result 4).$$\frac{d^2 y}{dx^2} = -\frac{(x+1) \cdot \left(- \frac{y + 1}{x + 1} \right) - (y+1)}{(x+1)^2}$$$$ = -\frac{-(y+1) - (y+1)}{(x+1)^2}$$$$= \frac{2(y+1)}{(x+1)^2}$$
Pro Tip
When calculating implicit derivatives, and especially when calculating second derivatives implicitly, be careful to show clean, easy-to-read scratch work. Profs often dock points for bad figuring. One tip I recommend is to calculate complicated pieces of a bigger process in separate steps. For example, in the last problem, it was helpful to calculate $d/dx$ of $xy$ on its own and then plug back in to the "main story".

Change Over Time Relationships

The last task that we are sometimes asked to perform with an implicit operator is to take a common formula with pieces that each change over time and differentiate it with respect to time. This mechanic will be of paramount importance in the upcoming lesson on Related Rates », so it's worth taking a look at both now and again.Let's start with an situation we can visualize.Imagine we have a circle that is expanding, such that the radius of the circle is expanding at a certain rate. The question we want to answer is, how fast is the area of the circle changing at any given time?The answer lies in implicitly differentiating the formula for circle area with respect to the variable $t$ (time). Here's why: the formula for the area of a circle is$$A = \pi r^2$$What we seek to know is, what is $dA/dt$, which is the way to measure rate of change of area as time changes, using the language of derivatives. However, since $A$ depends on $r$, and since $r$ depends on $t$ (since we said the radius is growing over time and thus its size depends on what time you observe it), we must apply the Chain Rule to the derivative using the implicit operator approach.$$\frac{d}{dt} \, A = \frac{d}{dt} \, \pi r^2 $$$$\frac{dA}{dt} = \pi \left(2r\right) \, \frac{dr}{dt}$$If we ignore the context, what happens on the right side of the equation is not very different from prior implicit examples such as$$\frac{d}{dx} \, y^2 = 2y \, \frac{dy}{dx}$$When we differentiate with respect to time, the "implicit process" for each variable is similar to what we've been learning throughout this lesson. The major difference is that there are multiple variables that each depend on $t$. Mechanically this will not add anything new - it's just more detail to track.
Example 9A right triangle has leg measurements $x$ and $y$.If the each legs of the triangle expand over time at rates of $dx/dt$ and $dy/dt$ respectively, find an expression for $dr/dt$, the rate at which the hypotenuse is expanding over time.$\blacktriangleright$ First, write down the relationship between $x$, $y$, and $r$.$$x^2 + y^2 = r^2$$(1)The question is asking us to find the rate at which the hypotenuse is changing, which is $dr/dt$. First, apply the implicit $d/dt$ operator to both sides of equation 1.$$\frac{d}{dt} \, \big[x^2 + y^2\big] = \frac{d}{dt} \, r^2$$$$2x \, \frac{dx}{dt} + 2y \, \frac{dy}{dt} = 2r \, \frac{dr}{dt}$$(2)Note that each expression has a chain rule factor attached, since the chain rule applies to all three variables $x$, $y$, and $r$ (each variable depends on $t$ because if we wanted to know the measure of any one of them, we would need to know how much time has passed as the legs grow).Now, isolate the thing we want - $dr/dt$ - by dividing both sides by $2r$.$$\frac{dr}{dt} = \frac{2x \, \frac{dx}{dt} + 2y \, \frac{dy}{dt}}{2r}$$(3)Furthermore, if we really wanted to get an expression that solely had $x$ and $y$ in it, since those are the independent changing variables, we could, if we want, replace $r$ with $\sqrt{x^2 + y^2}$, since that's how the three are related via equation 1.$$\frac{dr}{dt} = \frac{2x \, \frac{dx}{dt} + 2y \, \frac{dy}{dt}}{2\sqrt{x^2 + y^2}}$$(4)That last step may not be needed, depending on the situation. We can certainly say that result 3 is a correct answer since we did what we were asked to do.

Unusual Suspects - Keeping Variables Straight

Profs will usually not try to confuse you on purpose, but it is worth noting that the variable we are using for the derivative is important. Recall that when we were first introduced to the differential notation for derivatives, we said that the variable in the denominator should be the variable in our expression. E.g.$$\frac{d}{dx} \big[x^2 + 3x - 2\big]$$When we do this, we say that we are taking a derivative "with respect to" that variable. This is inconsequential up to this lesson, since we only have been working with $x$ expressions. But what about $y=y(x)$? $x$ is still the underlying variable but $y$ is a variable that depends on $x$. Therefore, asking for the derivative of a $y$ expression with respect to $x$ and asking for the derivative of a $y$ expression with respect to $y$ are two different questions. For example:$$\frac{d}{dx} \, y^3 = 3y^2 \cdot \frac{dy}{dx}$$following with the above implicit derivative definition based on the Chain Rule. However,$$\frac{d}{dy} \, y^3 = 3y^2$$with no factor after it. This is because in the second case, we're asking how $y^3$ changes as $y$ changes, but in the first case, we're asking how $y^3$ changes as $x$ changes. The underlying basis for this again comes from the Chain Rule. Technically the Chain Rule applies to every derivative you ever take, but if, for example, you are taking $x$ derivatives of $x$ expressions, the would-be Chain Rule factor is really a $1$. It's similar to how we never bother to write a $1$ exponent, or write a $1$ coefficient. We don't say $x^1$, we say $x$. We don't say $1x$, we say $x$. Here's a clarifying example of what I mean.Technically, by the Chain Rule:$$\frac{d}{dx} \, \sin(x) = \cos(x) \, \frac{d}{dx} \, x$$(1)just like$$\frac{d}{dx} \, \sin(x^2) = \cos(x^2) \frac{d}{dx} \, x^2$$(2)However, in Equation 1, the derivative of $x$ is $1$. Therefore, $\frac{d}{dx} \, \sin(x) = \cos(x)$ - a result that we are familiar with.Understanding what I mean about keeping variables straight in this section will give you a better jump on the dreaded upcoming implicit applied word problems known as Related Rates ».
You Should Know
99% of the time homework and exams on Implicit Differentiation will ask you to do $\frac{d}{dx}$ problems, and you won't have to pay attention to what the variable is that you are taking the derivative with respect to. However, this is a useful idea to digest for the upcoming lesson on Related Rates, and even further ahead, for understanding relationships among variables in a multivariable setting (advanced multivariable calculus).

Put It To The Test

Whew! Now that we've seen the many ways in which we need to understand and use implicit differentiation, let's drill on some realistic practice with questions that are similar to what most professors would put on a test.Implicit Derivatives of ExpressionsFind the implicit derivative of each of the following expressions with respect to $x$. Assume all non $x$ variables somehow depend on $x$.
Example 10$$-4y$$
Show solution
$$\blacktriangleright \,\, \frac{d}{dx} \, -4y = -4 \, \frac{dy}{dx}$$
Example 11$$3y^3$$
Show solution
$$\blacktriangleright \,\, \frac{d}{dx} \, 3y^3 = 9y^2 \, \frac{dy}{dx}$$
Example 12$$x e^x$$
Show solution
$\blacktriangleright$ This one is an $x$ only expression. We will proceed the same way, but there will be no $y$ or $dy/dx$ terms along the way. It is a straight-forward product rule example from what we've learned in the past.$$\frac{d}{dx} \, x e^x = x \, \frac{d}{dx} e^x + \frac{d}{dx} \, x + e^x$$$$= x e^x + e^x$$$$=e^x (x + 1)$$
Example 13$$\sin(xy)$$
Show solution
$\blacktriangleright$ Following the chain rule:$$\frac{d}{dx} \, \sin(xy) = \cos(xy) \cdot \frac{d}{dx} [xy]$$$$=\cos(xy) \big[x \, \frac{dy}{dx} + y\big]$$
Example 14$$x^2 y + xy^2 + 3$$
Show solution
$$\blacktriangleright \,\, \frac{d}{dx} \, x^2 y + xy^2 + 3$$$$=\frac{d}{dx} \, x^2 y + \frac{d}{dx} \, xy^2 + \frac{d}{dx} \, 3$$This is a really good time to take my advice for complicated processes. Do each of these derivative calculations on their own as a side story, and then plug the results into the main story.$$\frac{d}{dx} x^2 y = 2xy + x^2 \, \frac{dy}{dx}$$$$\frac{d}{dx} xy^2 = y^2 + x \cdot 2y \frac{dy}{dx} = y^2 + 2xy \frac{dy}{dx}$$$$\frac{d}{dx} 3 = 0$$Therefore,$$\frac{d}{dx} \, x^2 y + xy^2 + 3$$$$= \left( 2xy + x^2 \, \frac{dy}{dx} \right) + \left(y^2 + 2xy \frac{dy}{dx} \right) + 0$$$$= 2xy + x^2 \, \frac{dy}{dx} + y^2 + 2xy \frac{dy}{dx}$$
Example 15$$x^2 y + z \sqrt{y}$$
Show solution
$$\blacktriangleright \,\, \frac{d}{dx} \, \left[ x^2 y + y^{1/2} z \right]$$Each of these two pieces requires the product rule. Calculate each piece separately to avoid confusion.$$\frac{d}{dx} \, x^2 y = 2xy + x^2 \, \frac{dy}{dx}$$$$\frac{d}{dx} \, y^{1/2} z = y^{1/2} \, \frac{dz}{dx} + \frac{1}{2} \cdot y^{-1/2} \, \frac{dy}{dx} \cdot z$$$$=\frac{dz}{dx} \, \sqrt{xy} + \frac{z}{2\sqrt{y}} \, \frac{dy}{dx}$$Putting these together, we have$$\frac{d}{dx} \big[ x^2 y + z \sqrt{y} \big]$$$$= 2xy + x^2 \, \frac{dy}{dx} + \frac{dz}{dx} \, \sqrt{xy} + \frac{z}{2\sqrt{y}} \, \frac{dy}{dx}$$
Solving for dy/dxSimilar to many of the introductory examples of this lesson, these usual suspects will present you with some variable expression (usually with $y$) that is a function of $x$, and ask you to compute $dy/dx$. As mentioned above, your job is to apply the differential operator and ultimately isolate the term $dy/dx$ as if it was a variable.Determine an expression for $dy/dx$ for the curve described by each of the following equations.
Example 16$$3x + y^4 = 5$$
Show solution
$\blacktriangleright$ First, apply the $d/dx$ operator to both sides of the equation. Remember that this is always the first step in an implicit problem.$$\frac{d}{dx} \, 3x + y^4 = \frac{d}{dx} \, 5$$The derivative of $3x$ with respect to $x$ is $3$ (this is typical). The derivative of $y^4$ with respect to $x$ is $4y^3 \, \frac{dy}{dx}$ due to the Chain Rule related concept mentioned earlier (this is new knowledge for this lesson). Finally, the derivative of $5$ with respect to $x$ is 0 (this is typical from prior knowledge). Therefore, we have$$\frac{d}{dx} \, 3x + y^4 = \frac{d}{dx} \, 5$$$$\longrightarrow 3 + 4y^3 \, \frac{dy}{dx} = 0$$$$4y^3 \, \frac{dy}{dx} = -3$$$$\frac{dy}{dx} = \frac{-3}{4y^3}$$Notice once again how we 1) apply the $d/dx$ differential operator to both sides like a multipier and 2) treat $dy/dx$ like a variable and isolate it. These are both steps you'll do in every implicit problem where you're asked to find $dy/dx$.
Example 17$$e^y + 3x^4 = y^2$$
Show solution
$\blacktriangleright$ Once again, apply the $\frac{d}{dx}$ operator to both sides of the equation.$$\frac{d}{dx} \, \big[ e^y + 3x^4 \big] = \frac{d}{dx} \, y^2$$$$e^y \, \frac{dy}{dx} + 12x^3 = 2y \, \frac{dy}{dx}$$Now, move all the $dy/dx$ terms to one side.$$12x^3 = 2y \, \frac{dy}{dx} - e^y \, \frac{dy}{dx}$$Factor out $dy/dx$ like it was a variable.$$12x^3 = \frac{dy}{dx} \bigg[ 2y - e^y \bigg]$$Finally, algebraically isolate $dy/dx$.$$\frac{12x^3}{2y - e^y} = \frac{dy}{dx}$$
Example 18$$x^3 + y^3 = 6xy$$
Show solution
$$\blacktriangleright \,\, \frac{d}{dx} \, \big[x^3 + y^3 \big] = \frac{d}{dx} \, 6xy$$$$3x^2 + 3y^2 \, \frac{dy}{dx} = 6x \frac{dy}{dx} + 6y$$Move all the $dy/dx$ terms to one side and solve for it.$$3x^2 - 6y = 6x \frac{dy}{dx} - 3y^2 \, \frac{dy}{dx}$$$$3x^2 - 6y = \frac{dy}{dx} \big[ 6x - 3y^2 \big]$$$$\frac{dy}{dx} = \frac{3x^2 - 6y}{6x - 3y^2} = \frac{x^2 - 2y}{2x - y^2}$$
Example 19$$y^4 + x^2y^2 = 4 + x^3 y$$
Show solution
$$\blacktriangleright \,\, \frac{d}{dx} \, \big[ y^4 + x^2y^2 \big] = \frac{d}{dx} \, \big[ 4 + x^3 y \big]$$Do each of these derivatives separately since some of them are fairly involved.$$\frac{d}{dx} \, y^4 = 4y^3$$$$\frac{d}{dx} \, x^2 y^2 = 2x\left(y^2\right) + x^2 \bigg(2y \, \frac{dy}{dx} \bigg)$$$$=2xy^2 + 2x^2 y \, \frac{dy}{dx}$$$$\frac{d}{dx} \, 4 = 0$$$$\frac{d}{dx} \, x^3 y = x^3 \, \frac{dy}{dx} + 3x^2 y$$Now, plug in and solve from $dy/dx$.$$4y^3 + 2xy^2 + 2x^2 y \, \frac{dy}{dx} = x^3 \, \frac{dy}{dx} + 3x^2 y$$$$4y^3 + 2xy^2 - 3x^2 y = x^3 \, \frac{dy}{dx} - 2x^2 y \, \frac{dy}{dx}$$$$4y^3 + 2xy^2 - 3x^2 y = \frac{dy}{dx} \bigg[ x^3 - 2x^2 y \bigg]$$$$\frac{dy}{dx} = \frac{4y^3 + 2xy^2 - 3x^2 y}{x^3 - 2x^2 y}$$
Tangent SlopesFind the tangent slope ($dy/dx$) of the curve described by each equation at the specified point.
Example 20$$3xy - y^2 = -10$$at the point $(1,-2)$
Show solution
$\blacktriangleright$ We will solve for $dy/dx$ and then plug in the known values of each $x$ and $y$.$$\frac{d}{dx} \, \big[3xy - y^2 \big] = \frac{d}{dx} \, -10$$$$3x \, \frac{dy}{dx} + 3y - 2y \, \frac{dy}{dx} = 0$$$$3x \, \frac{dy}{dx} - 2y \, \frac{dy}{dx} = -3y$$$$\frac{dy}{dx} \, \big[ 3x - 2y \big] = -3y$$$$\frac{dy}{dx} = \frac{-3y}{3x-2y}$$$$\frac{dy}{dx} \Bigg|^{(1,-2)} = \frac{-3(-2)}{3(1)-2(-2)} = \frac{6}{7}$$
Example 21$$x^3 + 2xy + y^2 = 4$$at the point $(1,1)$
Show solution
$$\blacktriangleright \,\, \frac{d}{dx} \, \big[x^3 + 2xy + y^2 \big] = \frac{d}{dx} \, 4$$$$3x^2 + \big[ 2x \, \frac{dy}{dx} + 2y \big] + 2y \, \frac{dy}{dx} = 0 $$$$2x \, \frac{dy}{dx} + 2y \, \frac{dy}{dx} = -3x^2 - 2y$$$$\frac{dy}{dx} \, \big[ 2x + 2y \big] = -3x^2 - 2y$$$$\frac{dy}{dx} = \frac{-3x^2 - 2y}{2x + 2y}$$$$\frac{dy}{dx} \Bigg|^{(1,1)} = \frac{-3 - 2}{2 + 2} = \frac{-5}{4}$$
Example 22$$x^2y^2 - 2x = 3$$at the point $(3,-1)$
Show solution
$$\blacktriangleright \,\, \frac{d}{dx} \, \big[ x^2y^2 - 2x \big] = \frac{d}{dx} \, 3$$$$2x y^2 + x^2 (2y) \, \frac{dy}{dx} - 2 = 0$$$$2x^2 y \, \frac{dy}{dx} = 2 - 2xy^2$$$$\frac{dy}{dx} = \frac{2 - 2xy^2}{2x^2 y}$$$$\frac{dy}{dx}\Bigg|^{(3,-1)} = \frac{2 - 2(3)(-1)^2}{2(3)^2 (-1)} = \frac{-4}{-18}$$$$=\frac{2}{9}$$
Higher Implicit DerivativesFind each of the specified derivative expressions.
Example 23For the equation$$x^2 + y^2 = 16$$find $\frac{d^2 y}{dx^2}$
Show solution
$\blacktriangleright$ First find $dy/dx$.$$\frac{d}{dx} \, \big[ x^2 + y^2 \big] = \frac{d}{dx} \, 16$$$$2x + 2y \, \frac{dy}{dx} = 0$$$$\Rightarrow \frac{dy}{dx} = \frac{-x}{y}$$Now apply the $d/dx$ operator to both sides of this result:$$\frac{d}{dx} \, \frac{dy}{dx} = \frac{d}{dx} \, \frac{-x}{y}$$Use the quotient rule on the right side:$$\frac{d^2 y}{dx^2} = \frac{y(-1) - (-x) \, \frac{dy}{dx}}{y^2}$$$$=\frac{x \, \frac{dy}{dx} - y}{y^2}$$Replace the expression $dy/dx$ with the expression $-x/y$ since that's what we had previously figured out $dy/dx$ to be:$$\frac{d^2 y}{dx^2} = \frac{x\, \left( \frac{-x}{y}\right) - y}{y^2}$$$$=\frac{\frac{-x^2}{y}-y}{y^2}$$$$=\frac{-\frac{x^2}{y} - \frac{y^2}{y}}{y^2} = \frac{-x^2 - y^2}{y^3}$$
Example 24For the equation$$\sqrt{x} + \sqrt{y} = 1$$find $\frac{d^2 y}{dx^2}$
Show solution
$$\blacktriangleright \,\, \frac{d}{dx} \, \big[ x^{1/2} + y^{1/2} \big] = \frac{d}{dx} \, 1$$$$\frac{1}{2} \, x^{-1/2} + \frac{1}{2} \, y^{-1/2} \, \frac{dy}{dx} = 0$$$$\frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}} \, \frac{dy}{dx} = 0$$$$\frac{1}{2\sqrt{y}} \, \frac{dy}{dx} = -\frac{1}{2\sqrt{x}}$$$$\frac{dy}{dx} = -\frac{2\sqrt{y}}{2\sqrt{x}} = -\frac{\sqrt{y}}{\sqrt{x}}$$Now that we have determined the first derivative, apply the $d/dx$ operator to each side to find the second derivative.$$\frac{d}{dx} \, \frac{dy}{dx} = \frac{d}{dx} \, -\frac{\sqrt{y}}{\sqrt{x}}$$$$\frac{d^2 y}{dx^2} = - \frac{\sqrt{x} \, \frac{d}{dx} \, \sqrt{y} - \sqrt{y} \, \frac{d}{dx} \, \sqrt{x}}{\big( \sqrt{x} \big)^2}$$$$=-\frac{\sqrt{x} \left(\frac{1}{2\sqrt{y}}\right) \, \frac{dy}{dx} - \sqrt{y} \left( \frac{1}{2\sqrt{x}} \right)}{x}$$As before, replace the $dy/dx$ expression with its equivalent from earlier in the problem:$$=-\frac{\sqrt{x} \, \left(\frac{1}{2\sqrt{y}}\right) \left( -\frac{\sqrt{y}}{\sqrt{x}} \right) - \frac{\sqrt{y}}{2\sqrt{x}}}{x}$$$$=\Bigg[ \frac{\sqrt{y}}{2\sqrt{x}} + \frac{1}{2} \Bigg] \cdot \frac{1}{x}$$$$=\frac{\sqrt{y} + \sqrt{x}}{2x\sqrt{x}}$$
Example 25For the equation$$x^3 - y^3 = 10$$find $\frac{d^3 y}{dx^3}$
Show solution
$\blacktriangleright$ This is not commonly asked for, but if your prof is in a bad mood it can be. As you might expect, the protocol here is to solve for the second derivative, and then once again apply the $d/dx$ operator to that result to obtain the third derivative.$$\frac{d}{dx} \, \big[ x^3 - y^3 \big]= \frac{d}{dx} \, 10$$$$3x^2 - 3y^2 \, \frac{dy}{dx} = 0$$$$\frac{dy}{dx} = \frac{x^2}{y^2}$$There's the first derivative. Apply again to get the second.$$\frac{d}{dx} \, \frac{dy}{dx} = \frac{d}{dx} \, \frac{x^2}{y^2}$$$$\frac{d^2 y}{dx^2} = \frac{d}{dx} \, x^2 \cdot y^{-2}$$Rewriting it that way allows us to use the Product Rule instead of the Quotient Rule.$$\frac{d^2 y}{dx^2} = \left(x^2\right) \left( -2y^{-3} \right) \, \frac{dy}{dx} + (2x) \left(y^{-2}\right)$$$$=\frac{-2x^2}{y^3} \, \frac{dy}{dx} + \frac{2x}{y^2}$$As usual, substitute in for the $dy/dx$ expression we had found earlier ($x^2 / y^2$ in this case).$$\frac{d^2 y}{dx^2} = -\frac{2x^2}{y^3} \left( \frac{x^2}{y^2} \right) + \frac{2x}{y^2}$$$$=-\frac{2x^4}{y^5} + \frac{2x}{y^2}$$Now that we've found the second derivative, apply the $d/dx$ operator one last time to get the third derivative, simplifying as much as reasonably possible.$$\frac{d}{dx} \, \frac{d^2 y}{dx^2} = \frac{d}{dx} \, \bigg[ -\frac{2x^4}{y^5} + \frac{2x}{y^2} \bigg]$$$$\frac{d^3 y}{dx^3} = \frac{d}{dx} \, \big(-2x^4\big)\big(y^{-5}\big) + \frac{d}{dx} \, (2x)\big(y^{-2}\big)$$Again, third derivatives aren't frequently asked for and this is why. The calculations are detailed and complex. Let's calculate each of those derivatives separately. The terms on the right have already been rewritten so that we can apply the Product Rule instead of the Quotient Rule. Along the way we'll need to plug in the $dy/dx$ expression as well.$$\frac{d}{dx} \, \big(-2x^4\big)\big(y^{-5}\big) = -2x^4 \big(-5y^{-6}\big) \, \frac{dy}{dx} + \big(-8x^3\big)\big(y^{-5}\big)$$$$=-2x^4 \big(-5y^{-6}\big) \frac{x^2}{y^2} + \big(-8x^3\big)\big(y^{-5}\big)= \frac{10x^6}{y^8} - \frac{8x^3}{y^5}$$$$\frac{d}{dx} \, (2x)\big(y^{-2}\big) = 2x \big(-2y^{-3}\big) \, \frac{dy}{dx} + (2)\big(y^{-2}\big)$$$$=2x \big(-2y^{-3}\big) \left( \frac{x^2}{y^2} \right) + (2)\big(y^{-2}\big)$$$$=-\frac{4x^3}{y^5} + \frac{2}{y^2}$$Finally we can put these pieces in place and call it a day.$$\frac{d^3 y}{dx^3} = \frac{10x^6}{y^8} - \frac{8x^3}{y^5} -\frac{4x^3}{y^5} + \frac{2}{y^2}$$
Implicit Time DerivativesFind $d/dt$ of each of the following expressions.
Example 26$$V = \frac{\pi r^2 h}{3}$$
Show solution
$$\blacktriangleright \,\, \frac{d}{dt} \, V = \frac{d}{dt} \, \frac{\pi r^2 h}{3}$$The right side requires the product rule.$$\frac{dV}{dt} = \frac{\pi}{3} \Bigg[ 2r \, \frac{dr}{dt} \, h + r^2 \, \frac{dh}{dt} \Bigg]$$
Example 27$$S = 4\pi r^2$$
Show solution
$$\blacktriangleright \,\, \frac{d}{dt} \, S = \frac{d}{dt} \, 4\pi r^2$$$$\frac{dS}{dt} = 8 \pi r \, \frac{dr}{dt}$$
Example 28$$A = lw$$
Show solution
$$\blacktriangleright \,\, \frac{d}{dt} \, A = \frac{d}{dt} lw$$$$\frac{dA}{dt} = l \, \frac{dw}{dt} + w \, \frac{dl}{dt}$$
Example 29$$V = lwh$$
Show solution
$\blacktriangleright$ This one involves the Product Rule multiple times. If you're unsure how it works for a product of three objects, just group them into two objects to start.$$\frac{d}{dt} \, V = \frac{d}{dt} [(lw)h]$$$$\frac{dV}{dt} = h \cdot \frac{d}{dt} \, (lw) + lw \cdot \frac{d}{dt} \, h$$$$= h \cdot \left( l \, \frac{dw}{dt} + w \, \frac{dl}{dt} \right) + lw \frac{dh}{dt}$$$$= lh \, \frac{dw}{dt} + wh \, \frac{dl}{dt} + lw \, \frac{dh}{dt}$$Hopefully you can see the pattern - but memorizing the three-way Product Rule isn't very important because you can reinvent it so quickly.
Lesson Takeaways
  • Understand conceptually and (perhaps more importantly) mechanically how Implicit Differentiation works
  • Be able to solve for $dy/dx$ using Implicit Differentiation via the "derivative operator" approach
  • Find and interpret tangent slopes of curves that can only be quantified using implicit differentiation
  • Correctly apply the Product Rule as needed for Implicit Differentiation problems that involve the product of different variables
  • Know how to find higher derivatives with this process
  • Apply $d/dt$ to common formulas to understand how to set up rate of change over time formulas for related quantities

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