# Implicit Differentiation

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- 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

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 and worksheet coming soon!

## 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:## 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:- Take the implicit derivative of both sides of the equation by applying the derivative operator $\frac{d}{dx}$
- Take the derivative of each term with respect to $x$
- Move all the $dy/dx$ terms to one side of the equation, all other terms to the opposite side
- Isolate the $dy/dx$ term via algebra manipulation

## 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.## 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.## 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.## 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.## 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 ».## 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 $x$.- 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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