Implicit Differentiation of Parametric Equations

Posted on October 31, 2021 by Lin McMullin

I’ve never liked memorizing formulas. I would rather know where they came from or be able to tie it to something I already know. One of my least favorite formulas to remember and explain was the formula for the second derivative of a curve given in parametric form. No longer.

If $\displaystyle y=y(t)$ and, $\displaystyle x=x(t)$ then the traditional formulas give

$\displaystyle \frac{{dy}}{{dx}}=\frac{{dy/dt}}{{dx/dt}}$ , and

$\displaystyle \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}=\frac{{\frac{d}{{dt}}\left( {\frac{{dy}}{{dx}}} \right)}}{{\frac{{dx}}{{dt}}}}$

It is that last part, where you divide by $\displaystyle {\frac{{dx}}{{dt}}}$ , that bothers me. Where did the $\displaystyle {\frac{{dx}}{{dt}}}$ come from?

Then it occurred to me that dividing by $\displaystyle {\frac{{dx}}{{dt}}}$ is the same as multiplying by $\displaystyle {\frac{{dt}}{{dx}}}$

It’s just implicit differentiation!

Since $\displaystyle \frac{{dy}}{{dx}}$ is a function of t you must begin by differentiating the first derivative with respect to t. Then treating this as a typical Chain Rule situation and multiplying by $\displaystyle {\frac{{dt}}{{dx}}}$ gives the second derivative. (There is a technical requirement here that given $\displaystyle x=x(t)$ , then its inverse $\displaystyle t={{x}^{{-1}}}\left( x \right)$ exists.)

In fact, if you look at a proof of the formula for the first derivative, that’s what happens there as well:

$\displaystyle \frac{d}{{dx}}y(t)=\frac{{dy}}{{dt}}\cdot \frac{{dt}}{{dx}}=\frac{{dy/dt}}{{dx/dt}}$

The reason you do it this way is that since x is given as a function of t, it may be difficult to solve for t so you can find dt/dx in terms of x. But you don’t have to; just divide by dx/dt which you already know.

Here is an example for both derivatives.

Suppose that $\displaystyle x={{t}^{3}}-3$ and $\displaystyle y=\ln \left( t \right)$

Then $\displaystyle \frac{{dy}}{{dt}}=\frac{1}{t}$ and $\displaystyle \frac{{dx}}{{dt}}=3{{t}^{2}}$ and $\displaystyle \frac{{dt}}{{dx}}=\frac{1}{{3{{t}^{2}}}}$

Then $\displaystyle \frac{{dy}}{{dx}}=\frac{1}{t}\cdot \frac{{dt}}{{dx}}=\frac{1}{t}\cdot \frac{1}{{3{{t}^{2}}}}=\frac{1}{3}{{t}^{{-3}}}$

And $\displaystyle \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}=\left( {\frac{d}{{dt}}\left( {\frac{{dy}}{{dx}}} \right)} \right)\cdot \frac{{dt}}{{dx}}=\left( {-{{t}^{{-4}}}} \right)\cdot \left( {\frac{1}{{3{{t}^{2}}}}} \right)=-\frac{1}{{3{{t}^{6}}}}$

Yes, it’s the same thing as using the traditional formula, but now I’ll never have to worry about forgetting the formula or being unsure how to explain why you do it this way.

Revised: Correction to last equation 5/18/2014. Revised: 2/8/2016. Originally posted May 5, 2014.

Implicit Differentiation

Posted on September 28, 2017 by Lin McMullin

I discovered in doing next week’s post that I apparently never wrote about implicit differentiation. So here goes – an extra post this week!

Implicit Differentiation

The technique of implicit differentiation allows you to find slopes of relations given by equations that are not written as functions or may even be impossible to write as functions.

Example 1: A good way to start investigating this idea is to give your class the equation of a circle, say ${{x}^{2}}+{{y}^{2}}=25$ and ask them to find the slope of the tangent line (the derivative) where x = 3. No hints, just let them try.

Most students will hit upon solving for y and then differentiating:

$y=\pm \sqrt{{25-{{x}^{2}}}}$

$\displaystyle \frac{{dy}}{{dx}}=\frac{{-2x}}{{\pm 2\sqrt{{25-{{x}^{2}}}}}}=\frac{{-x}}{{\pm \sqrt{{25-{{x}^{2}}}}}}$

There are two points where x = 3: (3, 4) and (3, –4) at the first point the slope is – ¾ and at the second ¾.

Then show them another way – implicit differentiation.

To use this technique, assume that y is a function of x, but do not bother to find that function. Then using the chain rule on any terms containing a y. For ${{x}^{2}}+{{y}^{2}}=25$ , we have

$\displaystyle 2x+2y\frac{{dy}}{{dx}}=0$

Then solve for the derivative

$\displaystyle \frac{{dy}}{{dx}}=-\frac{x}{y}$

We see that this is the same as we found the first time, since $y=\pm \sqrt{{25-{{x}^{2}}}}$ ! There is a slight advantage here: we can now find the slopes from the coordinates without solving or dealing with the plus/minus sign. *

Example 2: Now let’s consider a more difficult example. Find the derivative of ${{x}^{2}}+4{{y}^{2}}=7+3xy$ . To solve for y here is possible but somewhat difficult (Hint: use the quadratic formula). We can continue writing ${y}'$ for dy/dx.

$2x+8y{y}'=0+3x{y}'+3y$

Note that the last term on the right is differentiated using the product rule. Since this happens fairly often, students need to be reminded of it.

Now solving for ${y}'$ gives

$\displaystyle {y}'=\frac{{3y-2x}}{{8y-3x}}$

Then we can find the derivatives at specific points by substituting the coordinates of the point. At the point (3,2) on the curve, the slope is $\displaystyle {y}'=\frac{{3\left( 2 \right)-2\left( 3 \right)}}{{8\left( 2 \right)-3\left( 3 \right)}}=\frac{0}{7}=0$

Note: the derivative of an implicit relation usually involves both the x and y coordinates.

Second Derivatives

This idea can be repeated to find second and higher derivatives.

Example 1 continued: In the first example with $\displaystyle \frac{{dy}}{{dx}}=\frac{{-x}}{y}$ we differentiate using the quotient rule:

$\displaystyle {{y}'}'=\frac{{y(-1)-(-x){y}'}}{{{{y}^{2}}}}$

The second derivative is a function, not just of x and y, but also of ${y}'$ . We can replace it with the first derivative and simplify.

$\displaystyle {{y}'}'=\frac{{-y+x\left( {\frac{{-x}}{y}} \right)}}{{{{y}^{2}}}}=\frac{{-{{y}^{2}}-{{x}^{2}}}}{{{{y}^{3}}}}=-\frac{{25}}{{{{y}^{3}}}}$

(This might be a good time to do a quick review of simplifying complex fractions; they occur often in implicit differentiation problems.)

To find the value of the second derivative at a given point we can substitute into either of the two expressions above. At (3, –4) where the derivative has been previously found to be ¾ we have $\displaystyle {{y}'}'=\frac{{y(-1)-(-x){y}'}}{{{{y}^{2}}}}=\frac{{\left( {-4} \right)\left( {-1} \right)-\left( {-3} \right)\left( {\frac{3}{{-4}}} \right)}}{{{{{\left( {-4} \right)}}^{2}}}}=-\frac{{-16-9}}{{-48}}=\frac{{25}}{{48}}$

Or we can use the second form

$\displaystyle {{y}'}'=-\frac{{25}}{{{{y}^{3}}}}=-\frac{{25}}{{{{{\left( {-4} \right)}}^{3}}}}=\frac{{25}}{{48}}$

Example 2 continued: The second example was taken from an AB Calculus exam (2004 AB 4). The first part gave the first derivative and asked students to show that it was correct. This was done (instead of just asking the students to find the first derivative) so that students would be sure to have the correct derivative to use later in the question.

The second part asked students to show that the tangent line is horizontal at the point where x = 3. This included finding the coordinates of the point, (3, 2) and showing that it is on the curve.

The third part of the question asked students to determine whether the point from part (b) was a relative maximum, a relative minimum or neither, and to justify their answer. Since there is no way to determine how the sign of the first derivative changes at the point the First Derivative Test cannot be used. Likewise, the Candidates’ Test (a/k/a the closed interval test) cannot be used without solving for y, and determining the domain of each part. That leaves the Second Derivative Test as the easiest choice.

$\displaystyle {y}'=\frac{{3y-2x}}{{8y-3x}}$ at (3,2) $\displaystyle {y}'=\frac{{3\left( 2 \right)-2\left( 3 \right)}}{{8\left( 2 \right)-3\left( 3 \right)}}=0$

$\displaystyle {{y}'}'=\frac{{\left( {8y-3x} \right)\left( {3{y}'-2} \right)-\left( {3y-2x} \right)\left( {8{y}'-3} \right)}}{{{{{\left( {8y-3x} \right)}}^{2}}}}$

Substituting the values into this without doing the algebra to remove the first derivative gives

$\displaystyle \begin{array}{l}{{y}'}'=\frac{{\left( {8\left( 2 \right)-3\left( 3 \right)} \right)\left( {0-2} \right)-\left( {3\left( 2 \right)-2\left( 3 \right)} \right)\left( {8\left( 0 \right)-3} \right)}}{{{{{\left( {8\left( 2 \right)-3\left( 3 \right)} \right)}}^{2}}}}=\frac{{\left( {16-9} \right)\left( {-2} \right)-0}}{{{{{\left( {16-9} \right)}}^{2}}}}=-\frac{2}{7}\\\end{array}$

So, the point (3, 2) is a relative maximum.

The graph of the relation, an ellipse is shown below.

* Incidentally, there is another clever way of doing example 1: The radius to any point on a circle centered at the origin has a slope of y/x. Since tangents to circles are perpendicular to the radii drawn to the point of tangency, the slope of the tangent must be –x/y.

Good Question 1: 2008 AB 6

Posted on January 21, 2015 by Lin McMullin

When I started this blog several years ago I was hoping my readers would ask questions that we could discuss or submit ideas for additional topics to write about. This has not really happened, but I’m still very open to the idea. (That was a HINT.) Since that first year when I had the entire curriculum ahead of me, I have written less not because I dislike writing, but because I am low on ideas.

The other day, I answered a question posted on the AP Calculus Community bulletin board about AB calculus exam question. It occurred to me that this somewhat innocuous looking question was quite good. So I decided to start an occasional series on good questions, from AP exams or elsewhere, that can be used to teaching beyond the actual things asked in the question. (My last post might be in this category, but that was written several months ago.)

In discussing these questions, I will make numerous comments about the question and how to take it further in your class. My idea is not just to show how to write a good answer, but rather to use the question to look deeper into the concepts involved.

Good Question #1: 2008 AB Calculus exam question 6.

The stem gave students the function $\displaystyle f\left( x \right)=\frac{\ln \left( x \right)}{x},\quad x>0$ . Students were also told that $\displaystyle {f}'\left( x \right)=\frac{1-\ln \left( x \right)}{{{x}^{2}}}$ .

The first thought that occurs is why they gave the derivative. The reason is, as we will see, that the first derivative is necessary to answer the first three parts of the question. Therefore, a student who calculates an incorrect derivative is going to be in big trouble (and the readers may have a great deal of work to do reading with the student’s incorrect work). The derivative is calculated using the quotient rule, and students will have to demonstrate their knowledge of the quotient rule later in this question; there is no reason to ask them to do the same thing twice.
If you are using this with a class, you can, and probably should, ask your students to calculate the first derivative. Then you can see how many giving the derivative would have helped.
When discussing the stem, you should also discuss the domain, x > 0, and the x-intercept (1, 0). Other features of the graph, such as end behavior, are developed later in the question, so they may be put on hold briefly.

Part a asked students to write an equation of the tangent line at x = e². To do this students need to do two calculations: $\displaystyle f\left( {{e}^{2}} \right)=\frac{2}{{{e}^{2}}}$ and $\displaystyle {f}'\left( {{e}^{2}} \right)=-\frac{1}{{{e}^{4}}}$ . An equation of the tangent line is $\displaystyle y=\frac{2}{{{e}^{2}}}-\frac{1}{{{e}^{4}}}\left( x-{{e}^{2}} \right)$ .

Writing the equation of a tangent line is a very important skill and should be straightforward. The point-slope form is the way to go. Avoid slope-intercept.
The tangent line is used to approximate the value of the function near the point of tangency; you can throw in an approximation computation here.
After doing part c, you should return here and discuss whether the approximation is an overestimate or an underestimate and how you can tell. (Answer: underestimate, since the graph is concave up here.)
After doing part c, you can also ask them to write the tangent line at the point of inflection and whether approximations near the point of inflection are overestimates or an underestimates, and why. (Answer: Since the concavity change here, it depends on which side of the point of inflection the approximation is made. To the left is an overestimate; to the right is an underestimate.)

Part b asked students to find the x-coordinate of the critical point, determine whether it is a maximum, a minimum, or neither, and to “justify your answer.” To earn credit students had to write the equation $\displaystyle {f}'\left( x \right)=0$ and solve it getting x = e. They had to state that this is a maximum because “ ${f}'\left( x \right)$ changes from positive to negative at x = e.”

This is a very standard AP exam question. To expand it in your class:

Discuss how you know the derivative changes sign here. This will get you into the properties of the natural logarithm function.
Discuss why the change in sign tells you this is a maximum. (A positive derivative indicates an increasing function, etc.)
After doing part c, you can return here and try the second derivative test.
The question asks for “the” critical point, hinting that there is only one. Students should learn to pick up on hints like this and be careful if their computation produces more or less than one.
At this point we have also determined that the function is increasing on the interval $\left( -\infty ,e \right]$ and decreasing everywhere else. The question does not ever ask this, but in class this is worth discussing as important features of the graph. On why these are half-open intervals look here.

Part c told students there was exactly one point of inflection and asked them to find its x-coordinate. To do this they had to use the quotient rule to find that $\displaystyle {{f}'}'\left( x \right)=\frac{-3+2\ln \left( x \right)}{{{x}^{3}}}$ , set this equal to zero and find the x-coordinate to be x = e^3/2.

The question did not require any justification for this answer. In class you should discuss what a justification would look like. The reason is that the second derivative changes sign here. So now you need to discuss how you know this.
Also, you can now determine that the function is concave down on the interval $\left( -\infty ,{{e}^{3/2}} \right)$ and concave up on the interval $\left({{e}^{3/2}},\infty \right)$ . Ask your class to justify this.

Part d asked student to find $\displaystyle \underset{x\to 0+}{\mathop{\lim }}\,\frac{\ln \left( x \right)}{x}$ . The answer is $-\infty$ . While this seems almost like a throwaway tacked on the end because they needed another point, it is the reason I like this question.

The question is easily solved: $\displaystyle \underset{x\to 0+}{\mathop{\lim }}\,\frac{\ln \left( x \right)}{x}=\underset{x\to 0+}{\mathop{\lim }}\,\frac{1}{x}\cdot \underset{x\to 0+}{\mathop{\lim }}\,\ln \left( x \right)=\left( \infty \right)\left( -\infty \right)=-\infty$ .
While tempting, the limit cannot be found by L’Hôpital’s Rule, because on substitution you get $\frac{-\infty }{0}$ ,which is not one of the forms that L’Hôpital’s Rule can handle.
The reason I like this part so much is that we have already developed enough information in the course of doing the problem to find this limit! The function is increasing and concave down on the interval $\left( -\infty ,e \right)$ . Moving from the maximum to the left, the function crosses the x-axis at (1, 0), keeps heading south, and gets steeper. So the limit as you approach the y-axis from the right is negative infinity.This is the left-side end behavior.
What about the right-side end behavior? (You ask your class.) Well, the function is positive and decreasing to the right of the maximum and becomes concave up after x = e^3/2. Thus, it must flatten out and approach the x-axis as an asymptote.
That $\displaystyle \underset{x\to \infty }{\mathop{\lim }}\,\frac{\ln \left( x \right)}{x}=0$ is clear from the note immediately above. This limit can be found by L’Hôpital’s Rule since it is an indeterminate of the type $\infty /\infty$ . So, $\displaystyle \underset{x\to \infty }{\mathop{\lim }}\,\frac{\ln \left( x \right)}{x}=\underset{x\to \infty }{\mathop{\lim }}\,\frac{\tfrac{1}{x}}{1}=0$ .
Notice also that the first derivative approaches zero as x approaches infinity. This indicates that the function’s graph approaches the horizontal as you travel farther to the right. The second derivative also approaches zero as x approaches infinity indicating that the function’s graph is becoming flatter (less concave).

This question and the discussion is largely done analytically (working with equations). We did find a few important numbers in the course of the work. Hopefully, you students discussed this with many good words. To complete the Rule of Four, here is the graph.

And here is a close up showing the important features of the graph and the corresponding points on the derivatives.

The function is shown in blue, the derivative and maximum in red, and the second derivative and the point of inflection in green.

Finally, this function and the limit at infinity is similar to the more pathological example discussed in the post of October 31, 2012 entitled Far Out!

Implicit Differentiation of Parametric Equations

Posted on May 17, 2014 by Lin McMullin

If $y=y\left( t \right)$ and, $x=x\left( t \right)$ then the tradition formula gives

$\displaystyle \frac{d}{dx}y\left( t \right)=\frac{dy/dt}{dx/dt}$ , and

$\displaystyle \frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{\frac{d}{dt}\left( \frac{dy}{dx} \right)}{\frac{dx}{dt}}$

It is that last part, where you divide by $\displaystyle \frac{dx}{dt}$ , that bothers me. Where did the $\displaystyle \frac{dx}{dt}$ come from?

Then it occurred to me that dividing by $\displaystyle \frac{dx}{dt}$ is the same as multiplying by $\displaystyle \frac{dt}{dx}$

It’s just implicit differentiation!

Since $\displaystyle \frac{dy}{dx}$ is a function of t you must begin by differentiating the first derivative with respect to t. Then treating this as a typical Chain Rule situation and multiplying by $\displaystyle \frac{dt}{dx}$ gives the second derivative. (There is a technical requirement here that given $x=x\left( t \right)$ , then $t={{x}^{-1}}\left( x \right)$ exists.)

In fact, if you look at a proof of the formula for the first derivative, that’s what happens there as well:

$\displaystyle \frac{d}{dx}y\left( t \right)=\frac{dy}{dt}\cdot \frac{dt}{dx}=\frac{dy/dt}{dx/dt}$

The reason you to do it this way is that since x is given as a function of t, it may be difficult to solve for t so you can find dt/dx in terms of x. But you don’t have to; just divide by dx/dt which you already know.

Here is an example for both derivatives.

Suppose that $x={{t}^{3}}-3$ and $y=\ln \left( t \right)$

Then $\displaystyle \frac{dy}{dt}=\frac{1}{t}$ , and $\displaystyle \frac{dx}{dt}=3{{t}^{2}}$ and $\displaystyle \frac{dt}{dx}=\frac{1}{3{{t}^{2}}}$ .

Then $\displaystyle \frac{dy}{dx}=\frac{1}{t}\cdot \frac{dt}{dx}=\frac{1}{t}\cdot \frac{1}{3{{t}^{2}}}=\frac{1}{3{{t}^{3}}}=\frac{1}{3}{{t}^{-3}}$

And $\displaystyle \frac{{{d}^{2}}y}{d{{x}^{2}}}=\left( \frac{d}{dt}\left( \frac{dy}{dx} \right) \right)\cdot \frac{dt}{dx}=\left( -{{t}^{-4}} \right)\cdot \frac{dt}{dx}=\left( -{{t}^{-4}} \right)\left( \frac{1}{3{{t}^{2}}} \right)=-\frac{1}{3{{t}^{6}}}$

Yes, it’s the same thing as using the traditional formula, but now I’ll never have to worry about forgetting the formula or being unsure how to explain why you do it this way.

Revised: Correction to last equation 5/18/2014

Revised: 2/8/2016

Teaching Calculus

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Tag Archives: second derivative

Implicit Differentiation of Parametric Equations

Implicit Differentiation

Good Question 1: 2008 AB 6

Implicit Differentiation of Parametric Equations

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