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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 1 (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)) at a point. Created by Sal Khan.

## Want to join the conversation?

- Isn't the distance from 2.5 and 4, 1.5 not 3?(7 votes)
- Definitely it is.

Sal actually is considering the boxes as distance from a point to point.

Since a box has a length of 0.5 units, the total distance is 1.5 units.(9 votes)

- Wouldn't it be easier to understand the slope of f(x) at 2.5 if Sal had explained it in terms of values of x and y instead of in terms of boxes on the graph? At3:40, Sal correctly stated,, "for every 3 that we run, we go up 2" and then he drew lines that went 3 BOXES to the right (from x=2.5 to x=4) and 2 BOXES up (from y=1 to y=2). I think some people were confused because going 3 boxes to the right is only going 1. 5 in terms of x and going 2 boxes up is only going up 1 in terms of y. The slope is indeed 2/3, but as Sal drew it was 1/1.5 (2-1/4-2.5) instead of 2/3.(4 votes)
- It really doesn't matter if you use 2/3 or 1/1.5. They both simplify to the same thing. Also, since the y-axis is proportional to the x-axis, all you really have to do is count boxes. As long as the proportion of each axis is equal, just counting boxes will give you the same answer as calculating the numbers involved. I hope this makes sense.(5 votes)

- So as long as the functions are continuous, or limited to a range where neither function change, is it possible to first solve for g(f(x)) = G(x) and then take the derivative like normal?

In this case, for the range 1 < x < 3, which includes x = 2.5, the functions look like:

f(x) = (2/3)x - 2/3

g(x) = 2x - 1

g(f(x)) = 2(*f(x)*) - 1

= 2((2/3)x - 2/3) - 1

= (4/3)x - 7/3

= G(x)

G(x) = (4/3)x - 7/3

Therefore G'(x) = 4/3 for all x, within the range 1 < x < 3

I also tried: G(x) = g(f(x))

f(x) = x^2

g(x) = 3x^2

and solved for G'(2)

Unless I did something wrong it seems to work here too. Is that correct? And does it hold true in general and with all functions?(3 votes)- As long as the functions are equivalent, their derivatives will be, so yes you can use this method. Sometimes it will be easier, but once you get the hang of the chain rule it becomes really easy to use.(3 votes)

- Is it possible to take the derivative of ANY graph, even if we don't have a mathematical definition such as x^2 of 7x^23+69x/5x? What if you have a graph with random variables or a graph made out of noise? Like the graphs here: https://www.khanacademy.org/computing/cs/programming-natural-simulations/programming-noise/a/perlin-noise

In the introduction Sal was talking about finding the instantaneous rate of change of runner. How do you that if a runner doesn't move in the kind of curve you can define by a function?(2 votes)- No, it is not possible to find a derivative for all functions at all points. Sometimes the derivative fails to exist even though the function is defined. Here are some major requirements for a function to be differentiable at some point x=c :

f(c) must be defined

f(x) must be continuous at x=c

lim h → 0⁺ [ f(x+h) - f(x) ] / h = lim h → 0⁻ [ f(x+h) - f(x) ] / h

(in other words, the left and right hand sides of the limit that is the definition of a derivative must be exactly equal)

If any of these is not met, the function is not differentiable at x=c

Random data or noise is not continuous and not necessarily even defined at the necessary points. Thus, it cannot be differentiated.(2 votes)

- Its given that G (x) = g (f (x) ) and G'(x) = g'(f (x) ) * f'(x). Lets suppose at some point X, the derivative is not defined for one of those functions...i.e. either g'(f (x) ) or f'(x) ....while the other is defined. So shall it be concluded that G'(x) wont be defined too ??(2 votes)
- how come f prime is 2/rds explain(1 vote)
- f prime of x Simply means the slope of f(x) at that value of x. So, in this case, the value of x was 2.5, and at 2.5, we can see that the slope is 2/3, meaning that at that point, for every 2 it goes vertically, it goes 3 horizontally.(1 vote)

- so since 1/1.5=1/(3/2)=2/3 Sal can use the box instead of the number since they all lead to the same rate of change?(1 vote)
- what if the graph of f(x) and g(x) are not a straight line, say a curve, what will happen to the f'(2.5) and g'(2.5)?(1 vote)
- It might be a bit harder to find the derivative, but the same process with chain rule would be used.(1 vote)

- dose it make a difference that he is using a scale that each tic is not 1, but .5?(1 vote)
- Nope. The numbers are still the same, just be careful to count each tic as .5, not one.(1 vote)

- Is the derivative of 2.5 not 0? Why?(1 vote)

## Video transcript

Consider the functions f and
g with the graphs shown below. If capital G of x is
equal to lowercase g of lowercase f of
x, what is the value of capital G prime of 2.5.? So G of x is a
composition of g and f. So it's g of f of x, or
lowercase g of f of x. And they don't graph
capital G of x here. They just give us the graphs of
lowercase g of x and lowercase f of x. This is the graph
of lowercase f of x. This is a graph of
lowercase g of x. So let's just try to think
how we could evaluate this and then see if they've given
us the right information here. So let me just rewrite a lot of
what they've already told us. They've already told
us that capital G of x is equal to lowercase
g of f of x. So if we wanted to take
the derivative of capital G of x-- and we do want
to think about what the derivative of capital G
of x is, because they want us to evaluate the derivative
at x is equal to 2.5. So let's do that. Let's take the derivative
of both sides of this. So if we take the derivative
of the left hand side, we end up with G prime of x. And on the derivative
on the right hand, since we have a composition
here of two functions, we would apply the chain rule. So this is going to be the
derivative of g with respect to f. So we could write
that as g prime of f of x times the derivative
of f with respect to x. So times f prime of x. So if we want to evaluate
what G prime of 2.5 is, then every place
we see an x here, we have to start
with 2.5 in there. So let's try to do that. So G prime of-- and I'll do this
in white so it sticks out-- G prime of 2.5 is going
to be equal to lowercase g prime of f of 2.5
times f prime of 2.5. So let's think about what
these would evaluate to. What is f of 2.5? Well, when x is
equal to 2.5-- let me just get a color
you can actually see. When x is equal to 2.5, our
function here is equal to 1. So f of 2.5, so we know
that f of 2.5 is equal to 1. Let me write that down,
f of 2.5 is equal to 1. And we also need to figure
out what f prime of 2.5 is. So f-- let me write it
this way-- f prime of 2.5 is equal to. Now what is f prime of 2.5? That's just
essentially the slope of the tangent line
at the function when x is equal to 2.5. So it's really just the
slope right over here. And at least right over at
this part of the function, it's actually a line. And the slope is actually
very easy to spot out. If we were to go from this
point to this point here-- and I'm just picking
those points because those are on kind of
integer-valued coordinates-- we see that for every three
that we run, we go up two, or that we rise two for
every three that we run. Or that our change in y
over change in x is 2/3. So the slope of the function
right over there is 2/3. So this is equal to 2/3. And so we can substitute back
in here, f of 2.5 is equal to 1. And this right over
here is equal to 2/3. Now, we're not done yet. Now we have to evaluate,
what is G prime of 1? So when x is equal to 1,
this is the function g. We're not evaluating g of 1. We're evaluating g prime of 1. So what is the slope
of the line here? Well, our change in y
over change in x is 2/1. If we go one in the
horizontal direction, we go up two in the
vertical direction. Change in y over
change in x is 2/1. So g prime-- let me write
this down-- g prime of 1 is equal to 2. So this whole thing
evaluates to 2. And so this simplifies--
let me scratch that out-- the simplifies to 2 times
2/3, which is equal to 4/3. So we could write G prime
of 2.5 is equal to 4/3. And this is a
pretty neat problem, because we didn't get to
see the actual function definition from G of x. But just using the chain rule
and the information they're giving us, we were
able to figure out what the value of
this derivative is when x is equal to 2.5.