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## AP®︎/College Calculus AB

### Course: AP®︎/College Calculus AB>Unit 3

Lesson 1: The chain rule: introduction

# Identifying composite functions

AP.CALC:
FUN‑3 (EU)
,
FUN‑3.C (LO)
,
FUN‑3.C.1 (EK)
Review of composite functions and how to recognize them. This is a valuable skill when approaching the Chain Rule in calculus.

## Want to join the conversation?

• Can't anything be considered a composite function?
• Well, yes, you can have u(x)=x and then you would have a composite function. In calculus, we should only use the chain rule when the function MUST be a composition. This is the only time where the chain rule is necessary, but you can use it whenever you want, technically.

Example - d/dx(3x+2). Clearly, the answer is 3, but we could use the chain rule:
u(x) = 3x
v(x) = x+2
d/dx(3x+2)=d/dx[v(u(x))])=1*3=3.
So the chain rule is usable but completely unnecessary.
• So at Sal said:
"So this would be cosine of u of x plus one." //where u of x is equal to sin(x)
But couldn't we say that u of x is equal to (sin(x) +1)?
Why did we leave the "1" out?
Or do we get to choose?
• At Sal states "I also want to state that there's oftentimes more than one way to compose ..."

So, yes you could choose to set u(x) = sin(x) + 1.

Try determining the derivative both the way Sal shows and this other way — you will find that you get the same answer.
• Hi... anything can be a composite function and chain rule can be applied but when is a function NECESSARILY a composite function?

• Well, a function is necessarily a composite function when you can't simplify it with known simplification techniques to arrive at a function comprised of a single function (a lot of functions have alternate definitions that you can use for simplification/complication later in your math education, like the link between exp(x) and trig functions). For example, (x^2)^2 isn't necessarily a composite function because with simplification, you can end up with x^4, which is a single function. Something that is necessarily a composite function is something like ln(sin(x)), which you can't simplify with elementary techniques to end up with a single function.
• say we have w(u(x)) we can take d/dx of this using the composite rule formula which it becomes
w'(u(x)) * u'(x)

But since this example starts talking about situations where you can use three functions at instead of two, how would you find the derivate then?

if g(x) = h(w(u(x))
does d/dx of g(x) become
= h'(w(u(x))) * w'(u(x)) * u'(x) ?

I did some working out and have seen that it can not and it seems that it becomes:
=h'(w(u(x))) * u'(x)

Without the middle term I had above w'(u(x))
I don't understand why I thought it should be in there. Is this correct? What is the proper way to think about the chain rule with more than 2 functions involved in the composition?
• You were right the first time, why did you decide it was: h'(w(u(x))) • u'(x)?

One way to remember this is to think of taking the derivative in steps:

Let f(x) = w(u(x)), so g(x) = h(f(x))

Apply the chain rule once:
g'(x) = h'(f(x)) • f'(x)

f'(x) = w'(u(x)) • u'(x)

Which gives us:
g'(x) = h'(w(u(x))) • w'(u(x)) • u'(x)
• Hey can't I make h(x)=Cos x and v(x)=sin x +1 then g(x)=h(v(x))=h(sin x +1)=cos(sin x +1) ??
• Yep, that looks about right. As a challenge, could you figure out what v(h(x)) would be?
• Sal says at that it would be hard to express sin(x)cos(x) as a composite function. What about:

Let f(x) = x*sin(cos^-1(x))

Let g(x) = cos(x)

Then f(g(x)) = cos(x)*sin(cos^-1(cos(x)) = cos(x)sin(x)
• You for sure can do that, but the only problem is, xsin(cos^-1 x) will not be defined for values greater than 1 or less than -1, which makes it "hard" to define it in some other way
• Easy tip.
In a composite function, one is INSIDE another.

Example: ln(sin(x)) because sin(x) is INSIDE ln(x),
except that we take the natural log of sin(x) instead of x in this composite function.
• can I apply the chain rule when I have a composition of three functions ? if it's possible, please write the formula for that.
• You can apply the chain rule to f(g(h(x))) by treating g(h(x)) as one function, so you get
f'(g(h(x))·[g(h(x))]'
Then evaluate the derivative of g(h(x)) by using chain rule again.