Calculus I Lesson 11: Implicit Differentiation
Review of Chain Rule
Recall the chain rule: if $h(x) = f(g(x))$, then $h^\prime(x) = f^\prime(g(x)) g^\prime(x)$.
Sometimes, this rule is thought of as a “substitution” rule. Let $y = f(g(x))$ and $u = g(x)$. Then we can think of $y$ as a function of $u$, as $y = f(u)$, or we can think of $y$ as a function of $x$, as $f(g(x)$. The chain rule relates the derivative of $y$ with respect to $u$, and the derivative of $y$ with respect to $x$:
\[\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\]The biggest challenge for using the chain rule is decomposing complicated functions into simpler functions that we “plug in” to each other. The textbook gives some good exercises on this: Section 3.6, #220-227. Let’s go through a couple of these:
221: $y = (3x^2 + 1)^3$. The “inside” function is $3x^2 + 1$, so let $u = 3x^2 + 1$. Then we substitute and $y = u^3$. So:
\[\frac{dy}{dx} = 3u^2 (\frac{du}{dx}) = 3(3x^2+1)^2 (6x)\]Or simplifying a little bit: $18x(3x^2+1)^2$.
222: $y = \sin^5(x)$. I usually would write this as $y = (\sin(x))^5$. So let $u = \sin(x)$ and $y = u^5$. Then $y^\prime = 5u^4 u^\prime$, or $5(\sin(x))^4 \cos(x)$.
On your own: go through the rest of the exercises 220 to 227. Ask in class or in office hours if you have trouble with them.
Implicit Differentiation
An important application of the Chain Rule is called “implicit differentiation”. Consider the following graph of the hyperbola $xy = 1$:
How might we find the equation of the tangent line at $x = 1$? The obvious way would be to solve for $y$ and then find the derivative of $\dfrac{1}{x}$. But we can also differentiate implicitly, meaning “take the derivative of an expression containing $y$”. That is, we can take the derivative of both sides of the equation $xy = 1$, and then use that to figure out what the derivative is when $x = 1$.
The derivative of $(xy)$ is, using the product rule, $x^\prime y + y^\prime x$. Since we know that the derivative of $x$ is $1$, this simplifies to $y + y^\prime x$. The derivative of $1$ is $0$, since that’s a constant. So if we take the derivative of both sides, we get:
\[y + y^\prime x = 0\]We can then solve for $y^\prime$ using algebra: subtract $y$ from both sides and then divide by $x$ to get $y^\prime = -\dfrac{y}{x}$.
In the original equation $xy = 1$, when $x = 1$, we can solve for $y$ and get $y = 1$ also. So we are looking for the equation of the line tangent to $xy = 1$ at the point $(1, 1)$. Plug in $x = 1$ and $y = 1$ to $y^\prime = -\dfrac{y}{x}$ and we get $y^\prime = -1$.
So our tangent line would have the equation $y - 1 = (-1)(x - 1)$, in point-slope form.
Take a look at this video from Khan Academy and look at how to find the derivatives of graphs of a circle.
Exercise: Find the equation of the line tangent to the graph of the unit circle $x^2 + y^2 = 1$ at the point $(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})$.
Exercises
Take a look at the exercises 300-309 from Section 3.8. Please make note of any of these that you have trouble with, and we can go over them next class.
Derivative of $y = \frac{1}{x}$
Earlier we used implicit differentiation to find equations of tangent lines to the graph of $xy = 1$. But we can use the same technique to find the (explicit) derivative of $y = \dfrac{1}{x}$. First, multiply by $x$ on both sides and then use implicit differentiation as before: $xy = 1$ means that $y + y^\prime x = 0$. Now solve for $y^\prime$, as we did before: $y^\prime = -\dfrac{y}{x}$.
This is what we got before, but is still an implicit equation. Instead, we can get an explicit equation, by substituting back in for $y$: since $y = \dfrac{1}{x}$, we get:
\[y^\prime = -\frac{\frac{1}{x}}{x} = -\frac{1}{x^2}\]We did a very similar thing earlier when we learned the quotient rule. Instead of taking the derivative of $y = \frac{f(x)}{g(x)}$, we took the derivative of $g(x)y = f(x)$ using the product rule, and then simplified and solved for $y^\prime$.
We will often use this method to help us find the derivative of inverse functions: if we have a function $y = f(x)$ and we know the derivative of $f(x)$, how might we find the derivative of its inverse $f^{-1}$? We use the fact that $y = f^{-1}(x)$ means the same thing as $x = f(y)$, and then differentiate implicitly.
Derivative of arctan
Here is an example of finding the derivative of an inverse function. $y = \arctan(x)$ means $x = \tan(y)$ (by definition: $\arctan$ is defined to be the inverse of $\tan$). Taking the derivative, implicitly, we get that $1 = (\sec(y))^2 y^\prime$. Since $\sec(y)$ means $\dfrac{1}{\cos(y)}$, we can multiply both sides by $(\cos(y))^2$ and get $y^\prime = (\cos(y))^2$. This is where things get tricky. How do we plug back in $y = \arctan(x)$ into $(\cos(y))^2$ and simplify? Let’s look at a triangle:
Let’s think about the relationship between the trig functions, inverse trig functions, and angles:
- Trig functions take angles as inputs, and output ratios (x / hypotenuse, y / hypotenuse, y / x, etc)
- Inverse trig functions take ratios as inputs, and output angles.
So if $y = \arctan(x)$, $y$ is the angle that we need to get $\tan(y) = x$ as our output. From the picture, if we want $\tan(\theta) = x$, we can draw a right triangle, with an angle $\theta$, whose opposite side is $x$ and adjacent side is $1$. Then the hypotenuse of the triangle (using the Pythagorean Theorem) is $\sqrt{x^2 + 1}$. In this case, we are looking for cosine of that angle, which would be $\dfrac{1}{\sqrt{x^2+1}}$. So:
\[y^\prime = (\cos(y))^2 = \left(\frac{1}{\sqrt{x^2+1}}\right)^2 = \frac{1}{x^2 + 1}\]Now we have an explicit derivative of $\arctan(x)$!
Review of Exponential Functions
An exponential function is a function of the form $y = b^x$, where $b$ is some constant. We exclude $b \leq 0$, since, for negative numbers, $b^{\frac{1}{2}}$ is not defined, for example.
Before we study the derivatives of these kinds of functions, I’d like to review these functions a bit. First of all, what do exponents even mean?
Exponents?
If that sounded like a rhetorical question, it wasn’t. We know what some exponents mean. For example, $2^3$ means $2 \times 2 \times 2$. We also know that $2^0$ is 1, and $2^{-4}$ means $\dfrac{1}{2 \times 2 \times 2 \times 2}$. So we can make sense of integer exponents.
We can also make sense of exponents that are fractions: $2^{\frac{3}{4}}$ means $\sqrt[4]{2^3}$. But what does $2^\pi$ mean?
Before we answer this, it’s worth mentioning: do we really know what the basic operations (addition, subtraction, multiplication, division) really mean for irrational numbers like $\pi$?
Addition? Multiplication?
We usually think of addition as repeated counting: $5 + 7 = 12$, because if we had $5$ apples on one side and $7$ apples on another, then we have $12$ apples total. But of course, there is no notion of $\sqrt{2}$ apples. What does $2 + \sqrt{2}$ mean? It’s really about measurement, rather than counting: “slide the number line over by a length of $\sqrt{2}$.”
Similarly, multiplication is usually thought of as repeated addition. $2 \times 3$ means $2$ groups of $3$. What does $\pi \times 3$ mean? $\pi$ groups of 3? Again, we need to re-think this notion slightly: instead of making groups, we re-scale: multiplying by $\pi$ means we stretch out the number line until the tick marks are spaced apart by a length of $\pi$.
In a sense our operations are transformations of the number line by sliding (addition) or stretching/shrinking (multiplication).
Incidentally, this helps us understand multiplying negatives: multiplying by a negative involves both the re-scaling as well as reflecting the whole line about the center. (So: what happens when you “reflect” the line once and then “reflect” it again? You get back where you started. Therefore: $-1 \times -1 = 1$!)
Exponents!
So what does this mean for exponents? The whole point is that we need to re-interpret exponentiation. Really, the only way to make sense of exponential functions is to use calculus: $2^\pi$ is defined as whatever value makes the function $2^x$ continuous as $x \rightarrow \pi$. In other words: we know what $2^3$ is, we know how to calculate $2^{3.1}$, and $2^{3.14}$, and $2^{3.141}$, and all of these numbers approach some limit as $x \rightarrow \pi$. $2^\pi$ is defined to be this limit.
Growth
Exponential functions are used to model continuous growth. Let’s go through a classical investment example.
Example: An investment earns 10% annually. We initially invest 1000 dollars. How much do we have after $t$ years?
Let $A(t)$ be the function that represents the amount of money invested after $t$ years.
- $A(0) = 1000$
- $A(1) = 1100$ (10% more than 1000: add 100)
- $A(2) = 1210$ (10% more than 1100: add 110)
- $A(3) = 1331$ (10% more than 1210: add 121)
This function is given, algebraically, by $A(t) = 1000 \cdot (1.1)^t$.
Notice that the amount the investment grows by is directly proportional to the amount invested. If we have 1,000,000 dollars invested, we will earn 100,000. If we have 1,000,000,000 invested, we will earn 100,000,000!
This is worth re-iterating: the key fact about exponential functions is that their growth is proportional to their value. In other words: if you divide the growth by the total, you should get the same constant each time (in the example above, it’s always 0.1). We will see that this fact is reflected in their derivatives: $y^\prime$ is proportional to $y$, no matter where we look on the graph; so $\frac{y^\prime}{y}$ will be a constant.
Next time we will study derivatives of exponential functions and their inverses, logarithmic functions. Please review Section 1.5 on exponential and logarithmic functions as well as Section 3.9 on the derivatives of exponential and logarithmic functions.
Reminders / Homework
- DeltaMath due Monday (10/13)
- Topic statement due following Monday (10/20)
- Presentation 1 starting next week.
- Written Homework 5 due next Thursday (10/16):
- Section 3.5: #180, 192, 201
- Section 3.6: #230, 232, 248, 254
- Section 3.8: #302, 316, 325