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

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

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?

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).

Why is it called the chain rule?

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

how do you know when to use chain rule?

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))

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!

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

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.

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.

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

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.

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

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.

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

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)

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

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.

Main content

AP®︎/College Calculus BC

Course: AP®︎/College Calculus BC > Unit 3

Lesson 1: The chain rule: introduction

Chain rule

Google Classroom

The chain rule tells us how to find the derivative of a composite function. Brush up on your knowledge of composite functions, and learn how to apply the chain rule correctly.

The chain rule says:

start fraction, d, divided by, d, x, end fraction, open bracket, f, left parenthesis, g, left parenthesis, x, right parenthesis, right parenthesis, close bracket, equals, f, prime, left parenthesis, g, left parenthesis, x, right parenthesis, right parenthesis, g, prime, left parenthesis, x, right parenthesis

It tells us how to differentiate composite functions.

Quick review of composite functions

A function is composite if you can write it as f, left parenthesis, g, left parenthesis, x, right parenthesis, right parenthesis. In other words, it is a function within a function, or a function of a function.

For example, start color #1fab54, cosine, left parenthesis, end color #1fab54, start color #e07d10, x, squared, end color #e07d10, start color #1fab54, right parenthesis, end color #1fab54 is composite, because if we let start color #1fab54, f, left parenthesis, x, right parenthesis, equals, cosine, left parenthesis, x, right parenthesis, end color #1fab54 and start color #e07d10, g, left parenthesis, x, right parenthesis, equals, x, squared, end color #e07d10, then start color #1fab54, cosine, left parenthesis, end color #1fab54, start color #e07d10, x, squared, end color #e07d10, start color #1fab54, right parenthesis, end color #1fab54, equals, start color #1fab54, f, left parenthesis, end color #1fab54, start color #e07d10, g, left parenthesis, x, right parenthesis, end color #e07d10, start color #1fab54, right parenthesis, end color #1fab54.

start color #e07d10, g, end color #e07d10 is the function within start color #1fab54, f, end color #1fab54, so we call start color #e07d10, g, end color #e07d10 the "inner" function and start color #1fab54, f, end color #1fab54 the "outer" function.

start color #1fab54, start underbrace, cosine, left parenthesis, space, start color #e07d10, start overbrace, x, squared, end overbrace, start superscript, start text, i, n, n, e, r, end text, end superscript, space, end color #e07d10, right parenthesis, end underbrace, start subscript, start text, o, u, t, e, r, end text, end subscript, end color #1fab54

On the other hand, cosine, left parenthesis, x, right parenthesis, dot, x, squared is not a composite function. It is the product of f, left parenthesis, x, right parenthesis, equals, cosine, left parenthesis, x, right parenthesis and g, left parenthesis, x, right parenthesis, equals, x, squared, but neither of the functions is within the other one.

Problem 1

Is g, left parenthesis, x, right parenthesis, equals, natural log, left parenthesis, sine, left parenthesis, x, right parenthesis, right parenthesis a composite function? If so, what are the "inner" and "outer" functions?

(Choice A)
g is composite. The "inner" function is natural log, left parenthesis, x, right parenthesis and the "outer" function is sine, left parenthesis, x, right parenthesis.
(Choice B)
g is composite. The "inner" function is sine, left parenthesis, x, right parenthesis and the "outer" function is natural log, left parenthesis, x, right parenthesis.
(Choice C)
g is not a composite function.

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

Usually, the only way to differentiate a composite function is using the chain rule. If we don't recognize that a function is composite and that the chain rule must be applied, we will not be able to differentiate correctly.

On the other hand, applying the chain rule on a function that isn't composite will also result in a wrong derivative.

Especially with transcendental functions (e.g., trigonometric and logarithmic functions), students often confuse compositions like natural log, left parenthesis, sine, left parenthesis, x, right parenthesis, right parenthesis with products like natural log, left parenthesis, x, right parenthesis, sine, left parenthesis, x, right parenthesis.

Problem 2

Is h, left parenthesis, x, right parenthesis, equals, cosine, squared, left parenthesis, x, right parenthesis a composite function? If so, what are the "inner" and "outer" functions?

(Choice A)
h is composite. The "inner" function is cosine, left parenthesis, x, right parenthesis and the "outer" function is x, squared.
(Choice B)
h is composite. The "inner" function is x, squared and the "outer" function is cosine, left parenthesis, x, right parenthesis.
(Choice C)
h is not a composite function.

Want more practice? Try this exercise.

Common mistake: Wrong identification of the inner and outer function

Even when a student recognized that a function is composite, they might get the inner and the outer functions wrong. This will surely end in a wrong derivative.

For example, in the composite function cosine, squared, left parenthesis, x, right parenthesis, the outer function is x, squared and the inner function is cosine, left parenthesis, x, right parenthesis. Students are often confused by this sort of function and think that cosine, left parenthesis, x, right parenthesis is the outer function.

Worked example of applying the chain rule

Let's see how the chain rule is applied by differentiating h, left parenthesis, x, right parenthesis, equals, left parenthesis, 5, minus, 6, x, right parenthesis, start superscript, 5, end superscript. Notice that h is a composite function:

\begin{aligned} h(x) &= \greenD{\underbrace{(~\goldD{\overbrace{5-6x}^{\text{inner}}~})^5}_{\text{outer}}} \\\\ \goldD{g(x)}&=\goldD{5-6x} &&\text{inner function} \\\\ \greenD{f(x)}&=\greenD{x^5}&&\text{outer function} \end{aligned}

Because h is composite, we can differentiate it using the chain rule:

start fraction, d, divided by, d, x, end fraction, open bracket, start color #1fab54, f, left parenthesis, end color #1fab54, start color #e07d10, g, left parenthesis, x, right parenthesis, end color #e07d10, start color #1fab54, right parenthesis, end color #1fab54, close bracket, equals, start color #11accd, f, prime, left parenthesis, start color #e07d10, g, left parenthesis, x, right parenthesis, end color #e07d10, right parenthesis, end color #11accd, dot, start color #ca337c, g, prime, left parenthesis, x, right parenthesis, end color #ca337c

Described verbally, the rule says that the derivative of the composite function is the inner function start color #e07d10, g, end color #e07d10 within the derivative of the outer function start color #11accd, f, prime, end color #11accd, multiplied by the derivative of the inner function start color #ca337c, g, prime, end color #ca337c.

Before applying the rule, let's find the derivatives of the inner and outer functions:

\begin{aligned} \maroonD{g'(x)}&=\maroonD{-6} \\\\ \blueD{f'(x)}&=\blueD{5x^4} \end{aligned}

Now let's apply the chain rule:

\begin{aligned} &\dfrac{d}{dx}\left[f\Bigl(g(x)\Bigr)\right] \\\\ =&\blueD{f'\Bigl(\goldD{g(x)}\Bigr)}\cdot\maroonD{g'(x)} \\\\ =&\blueD{5(\goldD{5-6x})^4} \cdot \maroonD{-6} \\\\ =&-30(5-6x)^4 \end{aligned}

Practice applying the chain rule

Problem 3.A

Current

Problem set 3 will walk you through the steps of differentiating sine, left parenthesis, 2, x, cubed, minus, 4, x, right parenthesis.

What are the inner and outer functions in sine, left parenthesis, 2, x, cubed, minus, 4, x, right parenthesis?

(Choice A)
The inner function is sine, left parenthesis, x, right parenthesis and the outer function is 2, x, cubed, minus, 4, x.
(Choice B)
The inner function is 2, x, cubed, minus, 4, x and the outer function is sine, left parenthesis, x, right parenthesis.
(Choice C)
The inner function is 2, x, cubed and the outer function is minus, 4, x.
(Choice D)
The inner function is minus, 4, x and the outer function is 2, x, cubed.

Problem 4

Current

start fraction, d, divided by, d, x, end fraction, open bracket, square root of, cosine, left parenthesis, x, right parenthesis, end square root, close bracket, equals, question mark

(Choice A)
minus, start fraction, cosine, left parenthesis, x, right parenthesis, divided by, 2, square root of, sine, left parenthesis, x, right parenthesis, end square root, end fraction
(Choice B)
start fraction, sine, left parenthesis, x, right parenthesis, divided by, 2, square root of, cosine, left parenthesis, x, right parenthesis, end square root, end fraction
(Choice C)
start fraction, cosine, left parenthesis, x, right parenthesis, divided by, 2, square root of, sine, left parenthesis, x, right parenthesis, end square root, end fraction
(Choice D)
minus, start fraction, sine, left parenthesis, x, right parenthesis, divided by, 2, square root of, cosine, left parenthesis, x, right parenthesis, end square root, end fraction

Want more practice? Try this exercise.

Problem 5

Current

x	f, left parenthesis, x, right parenthesis	h, left parenthesis, x, right parenthesis	f, prime, left parenthesis, x, right parenthesis	h, prime, left parenthesis, x, right parenthesis
minus, 1	9	minus, 1	minus, 5	minus, 6
2	3	minus, 1	1	6

G, left parenthesis, x, right parenthesis, equals, f, left parenthesis, h, left parenthesis, x, right parenthesis, right parenthesis

G, prime, left parenthesis, 2, right parenthesis, equals

Want more practice? Try this exercise.

Problem 6

Katy tried to find the derivative of left parenthesis, 2, x, squared, minus, 4, right parenthesis, cubed. Here is her work:

Step 1: Let start color #1fab54, f, left parenthesis, x, right parenthesis, equals, x, cubed, end color #1fab54 and start color #e07d10, g, left parenthesis, x, right parenthesis, equals, 2, x, squared, minus, 4, end color #e07d10, then start color #1fab54, left parenthesis, end color #1fab54, start color #e07d10, 2, x, squared, minus, 4, end color #e07d10, start color #1fab54, right parenthesis, cubed, end color #1fab54, equals, start color #1fab54, f, left parenthesis, end color #1fab54, start color #e07d10, g, left parenthesis, x, right parenthesis, end color #e07d10, start color #1fab54, right parenthesis, end color #1fab54.

Step 2: start color #11accd, f, prime, left parenthesis, x, right parenthesis, equals, 3, x, squared, end color #11accd

Step 3: The derivative is start color #11accd, f, prime, left parenthesis, end color #11accd, start color #e07d10, g, left parenthesis, x, right parenthesis, end color #e07d10, start color #11accd, right parenthesis, end color #11accd:

start fraction, d, divided by, d, x, end fraction, open bracket, start color #1fab54, left parenthesis, end color #1fab54, start color #e07d10, 2, x, squared, minus, 4, end color #e07d10, start color #1fab54, right parenthesis, cubed, end color #1fab54, close bracket, equals, start color #11accd, 3, left parenthesis, end color #11accd, start color #e07d10, 2, x, squared, minus, 4, end color #e07d10, start color #11accd, right parenthesis, squared, end color #11accd

Is Katy's work correct? If not, what's her mistake?

(Choice A)
Katy's work is correct.
(Choice B)
Step 1 is incorrect. left parenthesis, 2, x, squared, minus, 4, right parenthesis, cubed is equal to g, left parenthesis, f, left parenthesis, x, right parenthesis, right parenthesis, not f, left parenthesis, g, left parenthesis, x, right parenthesis, right parenthesis.
(Choice C)
Step 2 is incorrect. The derivative of f isn't 3, x, squared.
(Choice D)
Step 3 is incorrect. The derivative isn't f, prime, left parenthesis, g, left parenthesis, x, right parenthesis, right parenthesis.

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

A common mistake is for students to only differentiate the outer function, which results in f, prime, left parenthesis, g, left parenthesis, x, right parenthesis, right parenthesis, while the correct derivative is f, prime, left parenthesis, g, left parenthesis, x, right parenthesis, right parenthesis, g, prime, left parenthesis, x, right parenthesis.

Another common mistake: Computing f, prime, left parenthesis, g, prime, left parenthesis, x, right parenthesis, right parenthesis

Another common mistake is to differentiate f, left parenthesis, g, left parenthesis, x, right parenthesis, right parenthesis as the composition of the derivatives, f, prime, left parenthesis, g, prime, left parenthesis, x, right parenthesis, right parenthesis.

This is also incorrect. The function that should be inside of f, prime, left parenthesis, x, right parenthesis is g, left parenthesis, x, right parenthesis, not g, prime, left parenthesis, x, right parenthesis.

Remember: The derivative of start color #1fab54, f, left parenthesis, end color #1fab54, start color #e07d10, g, left parenthesis, x, right parenthesis, end color #e07d10, start color #1fab54, right parenthesis, end color #1fab54 is start color #11accd, f, prime, left parenthesis, end color #11accd, start color #e07d10, g, left parenthesis, x, right parenthesis, end color #e07d10, start color #11accd, right parenthesis, end color #11accd, start color #ca337c, g, prime, left parenthesis, x, right parenthesis, end color #ca337c. Not start color #11accd, f, prime, left parenthesis, end color #11accd, start color #e07d10, g, left parenthesis, x, right parenthesis, end color #e07d10, start color #11accd, right parenthesis, end color #11accd and not start color #11accd, f, prime, left parenthesis, end color #11accd, start color #ca337c, g, prime, left parenthesis, x, right parenthesis, end color #ca337c, start color #11accd, right parenthesis, end color #11accd.

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Anto Q
Posted 6 years ago. Direct link to Anto Q's post “Why are sin(x), cos(x), e...”
Why are sin(x), cos(x), etc, called transcendental functions?
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(36 votes)
Answer
- andrewp18
  Posted 6 years ago. Direct link to andrewp18's post “They cannot be expressed ...”
  They cannot be expressed as finite-degree polynomials.
  https://en.wikipedia.org/wiki/Transcendental_function
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  (64 votes)
jade jericho
Posted 5 years ago. Direct link to jade jericho's post “On Problem 2. I chose the...”
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?
Button navigates to signup pageComment on jade jericho's post “On Problem 2. I chose the...”
(28 votes)
Answer
- checedrodgers
  Posted 3 years ago. Direct link to checedrodgers's post “Yes, applying the chain r...”
  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).
  Button navigates to signup page
  (8 votes)
Hima Praveen
Posted 6 years ago. Direct link to Hima Praveen's post “how do you know when to u...”
how do you know when to use chain rule?
Button navigates to signup pageComment on Hima Praveen's post “how do you know when to u...”
(10 votes)
Answer
- Howard Bradley
  Posted 6 years ago. Direct link to Howard Bradley's post “One uses the chain rule w...”
  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))
  Button navigates to signup page
  (20 votes)
20leunge
Posted 6 years ago. Direct link to 20leunge's post “Why is it called the chai...”
Why is it called the chain rule?
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(11 votes)
Answer
- Moon Bears
  Posted 6 years ago. Direct link to Moon Bears's post “It is called the chain ru...”
  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
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  (10 votes)
Ryan
Posted 5 years ago. Direct link to Ryan's post “Why is the derivative of ...”
Why is the derivative of 3x = 3, but the derivative of 3(sin x) = 3(cos x)?
Button navigates to signup pageComment on Ryan's post “Why is the derivative of ...”
(4 votes)
Answer
- cossine
  Posted 5 years ago. Direct link to cossine's post “The derivative gives you ...”
  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.
  Button navigates to signup page
  (10 votes)
Liang
Posted 5 months ago. Direct link to Liang's post “can u not see [cos (x)]^2...”
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!
Button navigates to signup pageComment on Liang's post “can u not see [cos (x)]^2...”
(6 votes)
Answer
- Venkata
  Posted 5 months ago. Direct link to Venkata's post “Yep! You can do that. And...”
  Yep! You can do that. And as you said, you'll get the same answer in both cases (-2sin(x)cos(x))
  Comment on Venkata's post “Yep! You can do that. And...”
  (7 votes)
colinhill
Posted 3 months ago. Direct link to colinhill's post “One thing that has been a...”
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.
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(5 votes)
Answer
- kubleeka
  Posted 3 months ago. Direct link to kubleeka's post “That's the beauty of it: ...”
  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.
  Comment on kubleeka's post “That's the beauty of it: ...”
  (7 votes)
vishnuprashanth.cvi
Posted 6 years ago. Direct link to vishnuprashanth.cvi's post “Can someone kindly help m...”
Can someone kindly help me on how to differentiate (x-11)^3 / (x+3) using the chain rule?
TIA
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(1 vote)
Answer
- kubleeka
  Posted 6 years ago. Direct link to kubleeka's post “You'll need quotient or p...”
  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)
  Comment on kubleeka's post “You'll need quotient or p...”
  (11 votes)
Revant Sai T
Posted 4 years ago. Direct link to Revant Sai T's post “Can you please make a vid...”
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
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(2 votes)
Answer
- Alex
  Posted 4 years ago. Direct link to Alex's post “While the AM-GM inequalit...”
  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.
  Comment on Alex's post “While the AM-GM inequalit...”
  (5 votes)
hmc
Posted 5 years ago. Direct link to hmc's post “This is great. Is there a...”
This is great. Is there an article showing the chain rule when there are more than 2 functions in a composite?
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(0 votes)
Answer
- The #1 Pokemon Proponent
  Posted 5 years ago. Direct link to The #1 Pokemon Proponent's post “You can perfectly extend ...”
  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.
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  (9 votes)

AP®︎/College Calculus BC

Course: AP®︎/College Calculus BC > Unit 3

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))f'\big(g'(x)\big)f′(g′(x))f, prime, left parenthesis, g, prime, left parenthesis, x, right parenthesis, right parenthesis

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Another common mistake: Computing f, prime, left parenthesis, g, prime, left parenthesis, x, right parenthesis, right parenthesis