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Chain rule

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:
ddx[f(g(x))]=f(g(x))g(x)
It tells us how to differentiate composite functions.

Quick review of composite functions

A function is composite if you can write it as f(g(x)). In other words, it is a function within a function, or a function of a function.
For example, cos(x2) is composite, because if we let f(x)=cos(x) and g(x)=x2, then cos(x2)=f(g(x)).
g is the function within f, so we call g the "inner" function and f the "outer" function.
cos( x2inner )outer
On the other hand, cos(x)x2 is not a composite function. It is the product of f(x)=cos(x) and g(x)=x2, but neither of the functions is within the other one.
Problem 1
Is g(x)=ln(sin(x)) a composite function? If so, what are the "inner" and "outer" functions?
Choose 1 answer:

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 ln(sin(x)) with products like ln(x)sin(x).
Problem 2
Is h(x)=cos2(x) a composite function? If so, what are the "inner" and "outer" functions?
Choose 1 answer:

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 cos2(x), the outer function is x2 and the inner function is cos(x). Students are often confused by this sort of function and think that cos(x) is the outer function.

Worked example of applying the chain rule

Let's see how the chain rule is applied by differentiating h(x)=(56x)5. Notice that h is a composite function:
h(x)=( 56xinner )5outerg(x)=56xinner functionf(x)=x5outer function
Because h is composite, we can differentiate it using the chain rule:
ddx[f(g(x))]=f(g(x))g(x)
Described verbally, the rule says that the derivative of the composite function is the inner function g within the derivative of the outer function f, multiplied by the derivative of the inner function g.
Before applying the rule, let's find the derivatives of the inner and outer functions:
g(x)=6f(x)=5x4
Now let's apply the chain rule:
ddx[f(g(x))]=f(g(x))g(x)=5(56x)46=30(56x)4

Practice applying the chain rule

Problem 3.A
Problem set 3 will walk you through the steps of differentiating sin(2x34x).
What are the inner and outer functions in sin(2x34x)?
Choose 1 answer:

Problem 4
ddx[cos(x)]=?
Choose 1 answer:

Want more practice? Try this exercise.
Problem 5
xf(x)h(x)f(x)h(x)
19156
23116
G(x)=f(h(x))
G(2)=
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3/5
  • a simplified improper fraction, like 7/4
  • a mixed number, like 1 3/4
  • an exact decimal, like 0.75
  • a multiple of pi, like 12 pi or 2/3 pi

Want more practice? Try this exercise.
Problem 6
Katy tried to find the derivative of (2x24)3. Here is her work:
Step 1: Let f(x)=x3 and g(x)=2x24, then (2x24)3=f(g(x)).
Step 2: f(x)=3x2
Step 3: The derivative is f(g(x)):
ddx[(2x24)3]=3(2x24)2
Is Katy's work correct? If not, what's her mistake?
Choose 1 answer:

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(g(x)), while the correct derivative is f(g(x))g(x).

Another common mistake: Computing f(g(x))

Another common mistake is to differentiate f(g(x)) as the composition of the derivatives, f(g(x)).
This is also incorrect. The function that should be inside of f(x) is g(x), not g(x).
Remember: The derivative of f(g(x)) is f(g(x))g(x). Not f(g(x)) and not f(g(x)).

Want to join the conversation?

  • piceratops ultimate style avatar for user Anto Q
    Why are sin(x), cos(x), etc, called transcendental functions?
    (37 votes)
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  • blobby green style avatar for user jade jericho
    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?
    (34 votes)
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    • leaf green style avatar for user checedrodgers
      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).
      (15 votes)
  • piceratops seed style avatar for user Hima Praveen
    how do you know when to use chain rule?
    (10 votes)
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  • starky sapling style avatar for user 20leunge
    Why is it called the chain rule?
    (11 votes)
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  • aqualine ultimate style avatar for user Ryan
    Why is the derivative of 3x = 3, but the derivative of 3(sin x) = 3(cos x)?
    (4 votes)
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    • leaf green style avatar for user cossine
      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.
      (12 votes)
  • leafers seedling style avatar for user colinhill
    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.
    (6 votes)
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    • leaf green style avatar for user kubleeka
      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.
      (8 votes)
  • aqualine ultimate style avatar for user Liang
    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!
    (6 votes)
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  • blobby green style avatar for user vishnuprashanth.cvi
    Can someone kindly help me on how to differentiate (x-11)^3 / (x+3) using the chain rule?
    TIA
    (1 vote)
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    • leaf green style avatar for user kubleeka
      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)
      (12 votes)
  • starky tree style avatar for user hmc
    This is great. Is there an article showing the chain rule when there are more than 2 functions in a composite?
    (0 votes)
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  • female robot ada style avatar for user Revant Sai T
    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
    (2 votes)
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    • leaf grey style avatar for user Alex
      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.
      (5 votes)