Question 1

Why are sin(x), cos(x), etc, called transcendental functions?

Accepted Answer

They cannot be expressed as finite-degree polynomials.
https://en.wikipedia.org/wiki/Transcendental_function

Question 2

On Problem 2. I chose the answer "not a composite function" because i took cos^2(x) to be cos(x)*cos(x) and hence used the product rule to differentiate. Is that okay?

Accepted Answer

Yes, applying the chain rule and applying the product rule are both valid ways to take a derivative in Problem 2.

The placement of the problem on the page is a little misleading. Immediately before the problem, we read, "students often confuse compositions ... with products".

This suggests that the problem we are about to work (Problem 2) will teach us the difference between compositions and products, but, surprisingly, cos^2(x) is both a composition __and__ a product.

You can see this by plugging the following two lines into Wolfram Alpha (one at a time) and clicking "step-by-step-solution":
d/dx sin(x)cos(x)
d/dx cos(x)cos(x)

For d/dx sin(x)cos(x), W.A. applies the product rule. For d/dx cos(x)cos(x), W.A. recognizes that we can rewrite as a composition d/dx cos^2(x) and apply the chain rule.

In summary, there are some functions that can be written only as compositions, like d/dx ln(cos(x)). There are other functions that can be written only as products, like d/dx sin(x)cos(x). And there are other functions that can be written both as products and as compositions, like d/dx cos(x)cos(x).

Question 3

Why is it called the chain rule?

Accepted Answer

It is called the chain rule, because of the formula for the chain rule when considering differentiable functions from R^n to R^m. Read more at wikipedia https://en.wikipedia.org/wiki/Chain_rule#Higher_dimensions

Question 4

how do you know when to use chain rule?

Accepted Answer

One uses the chain rule when differentiating a function that can be expressed as a function of a function.
eg sin(x²) or ln(arctan(x))

Question 5

One thing that has been a little bit shaky with me is what can we define as its own function. I understand cos(x^2) being defined as f(x) = cos(x) and g(x) = x^2, but can we say cos(x) is also a composite function?

In that example,f(x) = cos(x) and g(x) = x then we solve with the chain rule. If we can, I'm assuming it's redundant, but I'm just looking for a "line" of when we can and can't define new functions. Also, can we combine different "instances" (sorry for the poor terminology) of variables as ONE unique function?

For example, if we had f(x) = (x^2)(cos(x))(sin(x)), could we combine (x^2) and cos(x) to be one function? so, g(x) = (x^2)(cos(x)) and h(x) = sin(x). If so, then could we apply the product rule with those two functions (one of which being a combination of both (x^2) and cos(x))? So, f(x) = (g(x))(h(x))

Sorry about the long question, I am just hoping for a more clear picture of when we can assign variables as their own function. Thanks.

Accepted Answer

That's the beauty of it: we can describe variables as separate functions however we like. We can say that x² is the function f(x)=x² composed with g(x)=x, and applying chain rule gives us
f'(g(x))·g'(x)
g'(x)=1, and g(x)=x, so this simplifies to
f'(x)
which is 2x.

Or we can rewrite x as e^(ln(x)). Then chain rule gives the derivative of x as
e^(ln(x))·(1/x), or x/x, or 1.

For your product rule example, yes we could consider x²cos(x) to be a single function, and in fact it would be convenient to do so, since we only know how to apply the product rule to products of two functions. By doing this, we find the derivative to be

d/dx[x²cos(x)]·sin(x)+(x²cos(x))·cos(x)

and now we can simplify this by computing the derivative of x²cos(x) using the product rule again.

There is no "line". We can divvy up expressions, introduce multiplications by 1, or write simple variables as compositions of inverse functions however we like, however makes the problem more convenient to solve.

Question 6

can u not see [cos (x)]^2 as a composite function, but see it as: cos (x)*cos (x), and use the product rule to find the derivative (using both chain rule and product rule ends up the same derivative)? thx!

Accepted Answer

Yep! You can do that. And as you said, you'll get the same answer in both cases (-2sin(x)cos(x))

Question 7

Why is the derivative of 3x = 3, but the derivative of 3(sin x) = 3(cos x)?

Accepted Answer

The derivative gives you a rate of change or to put it more simply the gradient of a function.

Since the rate change varies it is not the same. Like 3x =y, has a constant gradient of 3. But 3sinx has a gradient of 3cos(x) which depends on x which is not constant.

Question 8

Can you please make a video on how to find the minimum value of 2^sin(x) + 2^cos(x). Also, what is the inner function and outer function for 2^sin(x)??
 Thanks, any help would be appreciated

Accepted Answer

While the AM-GM inequality is useful, I'm just going to do it as a standard optimization problem.

Because you asked, inner function is sin(x) and outer function is 2^x.

f(x) = 2^(sin x) + 2^(cos x)
f'(x) = ln(2) * 2^(sin x) * cos x - ln(2) * 2^(cos x) * sin x
This is undefined nowhere, so we look for points where f'(x) = 0, or 2^(sin x) cos x = 2^(cos x) sin x; this expression is (essentially) equivalent to sin(x) = cos(x), which is true at pi/4 + k * pi.
The minimum value of f should be at 5pi/4, where (sin x, cos x) = (-sqrt(2) / 2, -sqrt(2) / 2).

min(f(x)) = f(5pi / 4 + 2pi * k) = 1.225.

Question 9

Can someone kindly help me on how to differentiate (x-11)^3 / (x+3) using the chain rule?
TIA

Accepted Answer

You'll need quotient or product rule in addition to the chain rule.

First let's find the derivative of (x-11)³. Outer function is x³, inner function is x-11. So we take
d/dx((x-11)³)
d/(x-11) (x-11)³ •d/dx (x-11)
3(x-11)²•1
3(x-11)²

Now, the derivative of x+3 is just 1. Putting these together with the quotient rule, we get
d/dx ((x-11)³/(x+3))
[(x+3)•d/dx(x-11)³ -(x-11)³•d/dx(x+3)]/((x+3)²)
Substitute the derivatives that we know and we get
[(x+3)•3(x-11)² -(x-11)³•1]/((x+3)²)
This is our answer. To simplify, we can factor an (x-11)² out of the numerator and get
(x-11)²•(3x+9 -(x-11))/((x+3)²)
(x-11)²•(2x+20)/((x+3)²)

Question 10

This is great. Is there an article showing the chain rule when there are more than 2 functions in a composite?

Accepted Answer

You can perfectly extend what applies for 2 functions to 3 functions or 10 functions or 100000 functions. For 2 functions (f(g(x)))'= f'(g(x))*g'(x) for 3 functions (f(g(h(x))))' = f'(g(h(x)))*g'(h(x)) *h'(x) and so on.

$x$ ‍	$f (x)$ ‍	$h (x)$ ‍	$f^{'} (x)$ ‍	$h^{'} (x)$ ‍
$- 1$ ‍	$9$ ‍	$- 1$ ‍	$- 5$ ‍	$- 6$ ‍
$2$ ‍	$3$ ‍	$- 1$ ‍	$1$ ‍	$6$ ‍

Course: AP®︎/College Calculus AB > Unit 3

Chain rule

Quick review of composite functions

Common mistake: Not recognizing whether a function is composite or not

Common mistake: Wrong identification of the inner and outer function

Worked example of applying the chain rule

Practice applying the chain rule

Common mistake: Forgetting to multiply by the derivative of the inner function

Another common mistake: Computing $f^{'} (g^{'} (x))$ ‍

Want to join the conversation?

AP®︎/College Calculus AB