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### Course: Calculus, all content (2017 edition) > Unit 2

Lesson 20: Chain rule- Chain rule
- Worked example: Derivative of cos³(x) using the chain rule
- Worked example: Derivative of ln(√x) using the chain rule
- Worked example: Derivative of √(3x²-x) using the chain rule
- Chain rule intro
- Chain rule overview
- Worked example: Chain rule with table
- Chain rule with tables
- Quotient rule from product & chain rules
- Chain rule with the power rule
- Applying the chain rule graphically 1 (old)
- Applying the chain rule graphically 2 (old)
- Applying the chain rule graphically 3 (old)
- Chain rule

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# Applying the chain rule graphically 3 (old)

Sal solves an old problem where the graphs of functions f and g are given, and he evaluate the derivative of [g(f(x))]² at a point. Created by Sal Khan.

## Want to join the conversation?

- No teacher or professor has ever been able to explain this to me! I've always heard that you can't treat differentials like numbers, but it always seems to work out that way (ex./ Derivation of Euler's Formula, the chain rule, implicit differentiation, etc.). What's the deal with that? Why shouldn't we think of them as "fractions" if we always end up treating them that way?(22 votes)
- You can't treat derivatives as true fractions because some of the terms are fixed and some of them are variable and you risk mixing them up. For example the derivative of log base b of x is the derivative of ln(x)/ln(b) (from logarithm properties, change of base formula). If you don't treat this as a derivative but as a fraction, you obtain the result b/x, because you'll take each derivative separately - as in deriv. of ln(x)=1/x and deriv. of ln(b)=1/b; (1/x)/(1/b)=b/x, which is wrong. Looking at Sal's example here (https://www.khanacademy.org/math/differential-calculus/taking-derivatives/chain_rule/v/derivative-of-log-with-arbitrary-base) you'll see that he takes out of the derivative the 1/ln(b), which is a constant term and the result is totally different, 1/(ln(b))x. The point here is that the ln(x) will vary with x, while the ln(b) has one and only one value.

Also to treat them as fractions would mean to miss a part of the derivative. For example, the deriv of e^cos(x), treated as a simple fraction would be just e^cos(x) because you would simplify the second part, which is incorrect. This derivative is e^cos(x) multiplied by the derivative of cos(x). Ths second part would be eliminated by simplification if you would treat derivatives as fractions.(15 votes)

- for along time i have been facing a problem in understanding the term something respect to something

if anybody can help I'll be pleased(3 votes)- The derivative of y(or any other variable)
**with respect to**x(or any other variable)**is the rate of change of the value of the variable 'y' with a extremely small change in the value of the variable 'x' or dy/dx**

The rate of change of distance with reference to time is called distance.

The rate of change of y can be found with any other variable other than x such as a which will represent the change in 'y' with a change in 'a'(14 votes)

- In which scenario would the derivative of a function be its negative?(4 votes)
- The simplest case: the zero function. If ƒ(x) = 0 for every x, then clearly ƒ '(x) = 0 = -ƒ(x).(3 votes)

- Can someone explain why the derivative of (sin(3x))^2 = 3sin(6x)?(2 votes)
- Not trying to be mean with redthumb, but he did a small mistake when he put that:

6•sin(2u) = 2•sin(u)•cos(u)

Just by comparing to the double-angle identity formula right above it we can see that the equation is wrong, thus leading to the incorrect result:

[3•sin(6•x)] = sin(3•x)•cos(3•x)

which is just the same as stating that:

sin(6•x) = (1 / 3)•sin(3•x)•cos(3•x)

which contradicts the double-angle identity formula.

Now what I'd do to apply that formula and get to the 3•sin(6x) answer is, after finding that:

d/dx(sin^2(3x)) = 6sin(3x)cos(3x)

sin(2θ) = 2sin(θ)cos(θ)

3x = θ

So now we have as a result:

6sin(θ)cos(θ)

Manipulating the DAI formula we get:

sin(θ)cos(θ) = sin(2θ) / 2

Substituting this in our result we have:

6sin(θ)cos(θ) = 6(sin(2θ) / 2) = (6 / 2)sin(2θ) = 3sin(2θ)

θ = 3x, so finally:

3sin(2•3x) = 3sin(6x)

Summarizing:

d/dx[sin²(3x)] = 6sin(3x)cos(3x) = 6[sin(2•3x) / 2] = 3sin(6x)(1 vote)

- What would g-prime of 5 have been? Since the slope of the line changes at that exact point.(3 votes)
- The derivative does not exist at that point.

There is no tangent line at that point.

The slope on the negative side of 5 does not equal the slope on the positive side of 5.(3 votes)

- what about the exponent 2 of the G(x)?? i think it won't lead the G'(5) to 0.(3 votes)
- @4:25you can see the fully expanded application of the chain rule on G(x). G'(x)= a product, and since one of the factors equals 0, the whole thing surely equals 0. Other than that, if the input on G(x) is constant, squaring it gives back a constant as well. When we graph G(x)=C on some interval, the slope of it is 0, and thus G'(x)=0 as well.(1 vote)

- What is the physical meaning of differentiating a function with respect to other function and how can we physically feel it and interpret graphically?

Like we can physically feel that what does differentiating a function with respect to x means (like we can graphically interpret; first principle of derivative).

Please help me with an understandable solution...

Thank You!(1 vote)- It's like peeling layers from an onion. You start with a degree 3 (cubic) onion and peel off the first layer in such a way that inside you find a second degree (square) onion. From the 3 dimensional onion you started with, after peeling off the first layer, you will obtain a bi-dimensional onion (the photo of the onion). Derivatives take things "one level down"; integrals, which are inverses of derivatives, take things "one level up". So if you start with the photo of an onion, by integrating it, you'll find the whole onion.

There is a difference between derivatives and differentials. Derivatives are taken for one variable while differentials are taken (somewhat circularly) with respect to all the variables. The derivative of an onion with respect to one variable will be the photo of the onion from one side. The differential of the onion can generate 2 views, one from side and one from (for example) the top of the onion. Of course, this can be generalized to an arbitrary number of dimensions, each with its own variable. The higher the degree of the onion (the higher dimensional the onion is), the more views you can generate by differentiating it.(4 votes)

- without using the triple derivative, would this process be right?

using the power rule and the chain rule at the same time,, basically.

G'(x)= 2(g(f(x))) * g'(f(x))* f'(x) ?(2 votes) - What can we use instead of f(x),g(x),h(x) and F(x),G(x),H(x)? Because we will need more function notations for differentiating a composition of more that 6 functions?(1 vote)
- You can use whatever symbol you want to designate your functions. So for starters you have all the letters in the alphabet (excluding those you design as variables), if you need more you can use greek letters, and if you need still more you can either decide to designate functions by a combination of letter and numbers (for example
`f₁(x)`

,`f₂(x)`

,`f₃(x)`

) or use abstract symbols or drawings to represent the functions.(2 votes)

- How do I use the
**dy/dx**notation? When I differentiate an expression, when should I use**d/du**and**d/dx**?(1 vote)- The "dy/dx" notation is used here to show the steps. When you differentiate using the chain rule, you won't need to write out the notation.

There are cases where you will use the dy/dx notation in the process of differentiation, but those will come later; namely implicit differentiation, where you will differentiate, then solve for dy/dx.(2 votes)

## Video transcript

Consider the functions f and
g with the graphs shown below. If capital G of x is equal to g
of f of x whole thing squared, what is the value of G
prime, capital G prime, of 5? And I encourage you to
now pause this video and try to solve it on your own. So let's try to think
through this somewhat complicated-looking function
definition right over here. So we have capital G of x. And actually, let me
write it this way. Let me write it this way,
I'll do it in yellow. We have capital G of x is
equal to this quantity squared. What we're squaring
is g of f of x. g of f of x is what
we're squaring. Or another way to
write G of x, If h of x were to be equal
to x squared, we could write G of x is equal
to h of this business, h of g of f of x. Let me just copy
and paste that so I don't have to keep
switching colors. So copy and paste, there we go. So this is another
way of writing G of x, where
whatever g of f of x, we input then to h of x, which
is really just squaring it. So there's a couple of
ways that we can write out the derivative of capital
G with respect to x. And you could
imagine this is going to involve the chain rule. But I like to write it
out, just to clarify in my head what's going
on and to make sure that it actually
makes some sense. So one thing that
we could write, we could write the
derivative of G with respect-- I'll mix
notations a little bit-- but I'll write the derivative
of G of x with respect to x is equal to the
derivative of this whole thing. So let me copy and paste
it, copy and paste. It's equal to this derivative
of this whole thing with respect to what's inside of
that whole thing. So if you wanted to treat
g of f of x as a variable, so with respect to that. So copy and paste. So it's going to be the
derivative of this whole thing with respect to g of f of
x times the derivative of g of f of x with respect to f
of x, with respect to-- I'll just copy and paste this
part, whoops-- with respect to f of x. And I like to write this out. It feels good. It looks like these are rational
expressions with differentials. It's really a notation more
than to be taken literally. But it feels good, or
at least in my mind it's a little bit more intuitive
why all of this works out. So with respect to f of x times
the derivative of-- and I'm using non-standard
notation here, but it helps me really
conceptualize this-- times the derivative of f of
x with respect to x. Or another way we
could write this is G prime of x is
equal to h prime of g of f of x, h
prime of-- actually, let me do it here--
h prime of this. So copy and paste,
h prime of that, times g prime of f of x,
times g prime of this. So copy and then paste. So times g prime of that. Put some parentheses there. Times f prime of x. And I like writing it this
way, because you notice if these were-- and once
again, this is more notation, but it gives a sense
of what's going on. If you did view
these as fractions, that would cancel with that. That would cancel with that. You're taking the derivative
of everything with respect to x, which is exactly
what you wanted to do. And let me put some
parentheses here so it makes a little bit
clearer what's going on. But this thing, in
my brain, I like to translate that
as, well, that's just h prime of g of f of x. This is g prime of f of x. This is f prime of x. And going from this
to try to answer your question, the question
that they're asking us actually isn't too bad. So we want to know,
what's G prime of 5? So everywhere we see an
x, let's change it to a 5. So we're going to say, we need
to figure out what G prime of 5 is. G prime of 5 is equal
to-- and actually, let me just copy and
paste this whole thing. So copy and paste. And so let me, everywhere
where I see an x, I'm going to
replace it with a 5. So let me get rid of that. Let me get rid of that. And let me get rid of that. And so I have a 5, a 5, and a 5. So what is f of 5? f of 5 is equal to negative 1. So this right over here
simplifies to negative 1. This right over here
simplifies to negative 1. And what's f prime
of negative 5? Well, that's the slope
of the tangent line at this point right over here. And we see that the
derivative, or the slope, of the tangent line here is 0. So this right over here
is going to be equal to 0. Now that's really interesting. So we could keep trying to,
well, what's g of negative 1? What's g prime of negative 1? You could see g of
negative 1, g of negative 1 we see is negative 1. g prime
of negative 1 is the slope here, which is also negative 1. Then we could calculate
h prime of these values, et cetera, et cetera. But we don't even
have to do that. Because this is the
product of three things, and one of these things
right over here is a 0. So 0 times anything
times anything is going to be equal to 0. Another way of thinking
about it is, f of x is isn't changing when
x is equal to 5. If f of x isn't changing
when x is equal to 5, then the input into the g
isn't going to be changing. So the g function isn't going
to be-- in the composition g of f of x-- isn't
going to be changing. And so h of g of f of x
isn't going to be changing. So g of x isn't
going to be changing. And so the derivative of
capital G of x at x equals 5 is going to be equal to 0.