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## Differentiation using multiple rules

# Product rule to find derivative of product of three functions

## Video transcript

What I want to do
in this video is think about how we can take
the derivative of an expression that can be viewed as a
product not of two functions but of three functions. And we're going to
do it using what we know of the product rule. And the way we
could think about it is we can view this as
the product, first, of two functions, of this
function here and then that function over there. And then separately
take the derivative. So if we just view the
standard product rule, it tells us that the
derivative of this thing will be equal to
the derivative of f of x-- let me close it with a
white bracket-- times the rest of the function. So times g of x--
let me close it with the-- times g of x
times h of x times plus just f of x times the
derivative of this thing. Times the derivative
with respect to x of g of x times h of x. Let me write that a
little bit neater. But what is this thing right
over here going to be equal? Well we can apply the
product rule again. So here, I'm just focusing
on this part right over here. The derivative of
this is just going to be g prime of x
times h of x plus g of x times the derivative
of h, times h prime of x. So everything that
we had the derivative of g of x times h of x is
this stuff right over here. And so we're going to
multiply that times f of x. So let's rewrite all
of this stuff here. So this first term right
over here we can rewrite. So all of this is going to be
equal to f prime of x-- that's that right over
there-- times g of x times h of x plus--
And now we're going to distribute this f of x. So it's f of x times this
plus f of x times this. So f of x times this is f
of x times g prime of x, the derivative of g, g
prime of x, times h of x. Let me do that in
that white color. And then finally,
f of x times this is just f of x times g
of x times h prime of x. And this was a
pretty neat result. Essentially, we can view
this as the product rule where we have three,
where we could have our expression viewed as
a product of three functions. Now we have three terms. In each of these terms, we
take a derivative of one of the functions and
not the other two. Here we took the
derivative of f. Here we took the
derivative of g. Here we took the
derivative of h. And you can imagine, if you had
the product of four functions here, you would have four terms. In each of them you'd be
taking a derivative of one of the functions. If you had n
functions here, then you would have n terms here. And in each of them you would
take the derivative of one of the functions. So this is kind
of a neat result.