[Instructor] Welcome to a proof of the Chain Rule
of Differentiation which is a rule used to find
the derivative of a composite function.
Let's first look at the Chain Rule using function notation
where the derivative of f of g of x with respect to x
is equal to f prime of g of x times g prime of x.
We often say the derivative of this composite function
is equal to the derivative of the outer function
evaluated at the inner function times the derivative
of the inner function where f is the outer function
and g is the inner function.
We could also express the Chain Rule using Leibniz notation
where if we let y equal f of g of x then d y d x
is equal to d y d u times d u d x, or in this case,
u would be the inner function, g of x.
So to begin our proof, we'll apply the limit definition
of the derivative so the derivative of f of g of x
with respect to x is equal to the limit as h approaches zero
of, looking at our definition below, for f of the quantity
x plus h, in our case we'd have f of g of the quantity
x plus h and then for minus f of x
we'd have minus f of g of x.
And of course, all this is divided by h.
Remember we obtained h by determining the difference
of the innermost inputs, so let's make a substitution for h,
let's replace h with the quantity x plus h
minus x, remember h would be the horizontal distance
between two points on the function f of g of x.
And now we're going to create an equivalent fraction
by multiplying this fraction by a fraction equal to one.
So for this proof we're going to multiply by this fraction here
which is equal to one as long as g of the quantity x plus h
doesn't equal g of x, otherwise we'd have
the indeterminate form of zero divided by zero.
So because we're multiplying by one this gives us
an equivalent fraction but instead of finding the products
we're going to use the commutative property of multiplication
and change the order of multiplication in the denominator.
So next you'll only change here is that notice that
g of the quantity x plus h minus g of x
is now the denominator of the first fraction
and the quantity x plus h minus x is now the denominator
of the second fraction, and again we can do this because
of the commutative property of multiplication.
Now we have the limit as h approaches zero of a product
because we have a limit of a product, we can write
this as a product of two limits.
So here we have the limit as h approaches zero
of the first fraction times the limit as h approaches zero
of the second fraction.
Now let's take this line to the next slide.
Let's take a look at the second limit first.
Notice how we could simplify the denominator
because we have the quantity x plus h minus x
which simplifies nicely to h,
and notice in this form we should recognize
this is the difference quotient for g of x
and therefore the limit as h approaches zero
of this difference quotient is going to be give us
g prime of x, but the reason I wanted to write
the denominator in this form is just to emphasize
this is the limit as h approaches zero of the change
in the outputs divided by the change in the inputs
where here we have an input of the quantity x plus h
and here we have an input of x,
and we can use this to help us understand how to determine
this first limit, where the limit is h approaches zero of,
this is a difference quotient but notice how our denominator
is g of the quantity x plus h minus g of x,
which if we look at the numerator,
these are the inputs into our function f.
Here we have f of g of the quantity x plus h
and here we have minus f of g of x.
So this is the difference quotient for f
but it's not the difference quotient for f of x.
It's the difference quotient for f of g of x.
And therefore the limit as h approaches zero
of this difference quotient gives us not f prime of x
but f prime of g of x and then we have times
the limit as h approaches zero of this difference quotient
for g of x, which gives us g prime of x,
and therefore we have our proof.
The derivative of f of g of x with respect to x is equal
to f prime of g of x times g prime of x.
We're using Leibniz notation, we would have d y d x equals
d y d u times d u d x.
I hope you found this helpful.