Chain rule example using visual function definitions | Differential Calculus | Khan Academy
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Differential calculus on Khan Academy: Limit introduction, squeeze theorem, and epsilon-delta definition of limits.
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Closed Caption:
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.
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