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I need to understand "implicit differentiation" and after that I need to be able to explain it to a student.

Here is an example:

Find the formula of a tangent line to the following curve at the given point using implicit differentiation.

x+xy+y^2=7 at a point (1,2)

What is the best way of explaining that?

Thank you.

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    $\begingroup$ Can you tell us what you're thinking so far? $\endgroup$
    – Sue VanHattum
    Sep 26 at 19:15
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    $\begingroup$ I suggest asking this question on math.stackexchange. I would be happy to answer there. $\endgroup$ Sep 26 at 20:32
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    $\begingroup$ Something I used to do, and I even discussed it briefly in a talk at the 2007 MathFest (see p. 65 here), is for short classroom quizzes I would include two graphs (using this software), one zoomed in a bit so students could estimate the slope as a safety check for their answers and the other zoomed out enough to give a sense of what the full graph looks like. $\endgroup$ Sep 26 at 20:33
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    $\begingroup$ When I teach implicit differentiation, I like to have students work through examples where the graph is given (as Dave suggests for quizzes). $\endgroup$
    – Sue VanHattum
    Sep 26 at 23:44
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Without being rigourous, the equation $F(x,y)=0$ defines a relation between $y$ and $x$. To promote this relation to a function, some restrictions in the $(x,y)$-plane has to be imposed. Assuming that such conditions exists, you can say that the equation $F(x,y)=0$ is defining $y$ implicitly as a function of $x$.

In some cases, we will be able to manipulate the equation $F(x,y)=0$ to find an explicit representation of the desired function, by solving the expression for $y$. In other cases, it won't be able to solve, but in both cases the derivate $y' = \dfrac{dy}{dx}$ should exists as a function.

A simple example:

$y^3 - x = 0$ defines $y$ implicitly as a function of $x$. But in this case, we can solve the expression for $y$, which leads to $y= \sqrt[3]{x}$, and then, $$\dfrac{dy}{dx} = \dfrac{1}{3\sqrt[3]{x^2}}$$ Implicit differentiation will help to compute the derivative without the solving-for-y process. This requires the chain rule, because in general: $$\dfrac{dL}{dx} = \dfrac{dL}{dy}\cdot \dfrac{dy}{dx}$$ Thus, using properties of derivatives,

$$y^3 - x = 0 \implies \dfrac{d(y^3)}{dx} - \dfrac{d(x)}{dx} = \dfrac{d(0)}{dx}$$ $$\dfrac{d(y^3)}{dy}\cdot \dfrac{dy}{dx} - 1 = 0$$ $$3y^2\cdot \dfrac{dy}{dx} - 1 = 0 \implies \dfrac{dy}{dx} = \dfrac{1}{3y^2}$$

It should be noted that $\dfrac{1}{3y^2}$ and $\dfrac{1}{3\sqrt[3]{x^2}}$ are indeed the same expression, provided the explicit representation $y=\sqrt[3]{x}$ of the function.

Another example

$e^{y}-y=x$ is an expression where there's no way to find an explicit representation. Nevertheless, you can assume that $y$ is a function of $x$ and use implicit differentiation to find the derivative of the function:

$$\dfrac{d(e^{y})}{dx} - \dfrac{d(y)}{dx} = \dfrac{d(x)}{x}$$ $$\dfrac{d(e^y)}{dy}\cdot \dfrac{dy}{dx} - \dfrac{dy}{dx} = 1 $$ $$\dfrac{dy}{dx} \left( e^y - 1\right) = 1 \implies \dfrac{dy}{dx} = \dfrac{1}{e^y-1}$$

enter image description here

As you can see, in the relation $e^{y}-y=x$ there are two ways (or "branches") to define $y$ as a function of $x$. The domain of the function is $[1,+\infty[$ and the range for the possitive branch is $[0,+\infty[$. This function is differentiable in $]1,+\infty[$, because for $x=1$ we have $y=0$ and therefore, the derivative is ill-defined. You can also deduce some other properties of the function. For example, since $y>0 \implies e^y>1$, the derivative is always possitive and therefore, the possitive branch is a stricly increasing function.

I hope this helps for the explanation.

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    $\begingroup$ Thank you for your explanation. $\endgroup$
    – Jan N.
    Sep 27 at 6:40
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I think it is important to constantly connect implicit differentiation with the chain rule. At first, there might be examples where a third variable is the independent variable, and $x$ and $y$ are dependent variables. For example, find $\frac{dy}{dx}$ if $x(t)^2+y(t)^2=1$, so $2x\frac{dx}{dt}+2y\frac{dy}{dt}=0$. In this case, the coordinate variables are of the same status: dependent variables. Somehow, putting the coordinate variables on this same status makes it more clear that the chain rule is in play. To finish, we do have to employ the chain rule again, in the form $\frac{dy}{dt}=\frac{dy}{dx}\cdot \frac{dx}{dt}.$

Then the next step is the situation were $x$ is the independent variable, and $y$ depends on $x$. So in effect, $x=t$ and $y=y(t)$ but we might as well drop $t$ and think $x=x$ and $y(x)$, and consider finding $\frac{dy}{dx}$ when $x^2+y^2=1$, but reinforce the notation $x^2+y(x)^2=1$. Students will be annoyed at having to write $x^2+y(x)^2=1$ but this is good medicine to enforce at first.

Eventually, students learn to first think about what variables are dependent and which is the independent variable, before jumping into the mechanics of differentiation rules. Only then can they be freed from the instructor's requirement of writing the explicit dependency such as $x(t)^2+y(t)^2=1$ or $x^2+y(x)^2=1$.

When I grade a problem like this on an exam, students know that they will at least get partial credit if they write the equation with the correct explicit dependency.

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  • $\begingroup$ I am intrigued by the temporary move to a parametric equation. IMHO, it will be a great method to use if we know the class has had a good experience with parametrics. It's taught in pre-calc, but results vary. For some the attitude is "I'm glad we got past that." For the other students, I think this approach is really useful. $\endgroup$ Sep 28 at 11:29
  • $\begingroup$ @JTP-ApologisetoMonica "Really" x and y are functions on the curve: they project points of the curve onto the x and y axis. The relation defining the curves is also a true equality of functions. Taking exterior derivatives is what "relates the rates". When the curve is parameterized locally (which is how we calculate everything with manifolds!) this becomes ordinary 1 variable calculus. $\endgroup$ Sep 28 at 15:25
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I haven't taught calculus in a while and looking back, implicit differentiation is one of the (many) topics that I think I could have taught better. If I was teaching it now I would emphasize that the derivative is a local property, and show how by zooming in on a point on the graph we can find a region where $y$ is a function of $x$ (except when we can't, and maybe the students can see when that happens). Let's call this function $g$.

Then, using your example, I would rewrite $x+xy+y^2=7$ as $x+xg(x)+g(x)^2=7$, and emphasize that it holds locally. Since $x+xg(x)+g(x)^2$ is a constant function in our little region, its derivative is zero there. Now we just use basic differentiation rules to get an expression for $g'$, and thus the tangent line.

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  • $\begingroup$ Completely agree. Too often implicit differentiation is taught as if one is manipulating the variables x and y. Naming the implicitly defined function, and emphasizing that it is a function (whether or not one can find a closed form algebraic expression for it) is essential. $\endgroup$ Sep 27 at 15:03
  • $\begingroup$ One can view x and y as projection functions from the curve to the axes, and then the manipulations make sense. However this is multivariable calculus (even differential geometry) and is (I think) too much for students to handle if you want them to actually have a logical view of the subject. $\endgroup$ Sep 27 at 15:05
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I introduce implicit differentiation in two parts:

  • How to differentiate dependent variables (i.e. $\frac{d}{dx} y = \frac{dy}{dx}$)
  • If two functions are equal not just at a point, their derivatives are equal

And then go through a bunch of examples including the squircle $x^4+y^4=1$ and plenty of just "find $\frac{d}{dx} \left(xy + x^2 + y^2\right)$"-type problems.

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    $\begingroup$ Might I suggest "If two functions are equal on an interval, then their derivatives are equal on that interval"? $\endgroup$ Sep 27 at 15:01

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