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## Calculus, all content (2017 edition)

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

Lesson 19: Product rule- Product rule
- Differentiating products
- Differentiate products
- Worked example: Product rule with table
- Worked example: Product rule with mixed implicit & explicit
- Product rule with tables
- Product rule to find derivative of product of three functions
- Product rule proof
- Product rule review

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# Product rule to find derivative of product of three functions

Sal differentiates the product of three different functions, and generalizes for the derivative of the product of any number of functions. Created by Sal Khan.

## Want to join the conversation?

- Whats the difference between "d/dx" and "dy/dx"? I see the notations can be used interchangeably in calculus, but when I took a peek in the differential equations section, there seemed to be a big difference between the two. Can someone please tell me what the difference is, if there is one?(41 votes)
- When finding the slope of a line, we think (deltaY/deltaX), where delta (∆) means "change in". Well, in the derivative operator, think about what the "d" means. "d" just means an infinitesimally small delta, or basically, as delta approaches 0. This is because we are trying to find the slope at a specific point, not at the segment between two points. Finding the slope of a segment between two points worked in basic algebra because the slope along a line is constant. Stay with me here.

So if we consider d/dx as an operator, it's really saying "the infinitesimally small change in f(x) over the infinitesimally small change in x", which is the derivative you are likely used to. You can probably see where this is going. dy/dx is just saying "the infinitesimally small change in y over the infinitesimally small change in x". So we use d/dx when we are talking about f(x), and we used dy/dx when we are talking about y.

So (d/dx)[x^2] = 2x, and (dy/dx)[y=x^2] = 2x.(101 votes)

- Does this work for any number of functions?(9 votes)
- The product rule will work for any number of functions though it will become quite complicated.(13 votes)

- Can someone give a proof of using product rule for n functions (n is any general number)?(3 votes)
- I apologize for the messiness of not being able to typeset that is about to ensue. Also, if you want an explanation of why a proof by induction works just let me know.

We want to show that the derivative d/dx (f_1 f_2 ... f_n) of n functions is equal to the sum from i=1 to n of (fi' * [product of f_j with j<= n and not equal to i]). So for example d/dx (f1 * f2 * f3)= f_1' f_2 f_3 + f_2' f_1 f_3 + f_3' f_1 f_2. We will prove this by induction.

Base case: This would just be a standard proof of the product rule for two functions.

Inductive hypothesis: We know assume that given k functions we know that d/dx (f_1 f_2 ... f_k) = sum from i=1 to k of (f_i' * [product of f_j with j<=k and not equal to i]). Using this fact we will also prove it is true for k+1 functions. To do this first group these k+1 functions like so: (f_1 f_2 ... f_k)(f_k+1). Well this is just the product of two functions so we can use the product rule to get d/dx (f_1 f_2 ... f_k)(f_k+1) = (f_1 f_2 ... f_k)' f_k+1 + f_k+1' (f_1 f_2 ... f_k). Our inductive hypothesis tells us that this must equal {sum from i=1 to k of (fi' * [product of f_j with j<=k and not equal to i])} f_k+1 + f_k+1' (f_1 f_2 ... f_k). We can multiply the f_k+1 through the sum to see this is equal to {sum from i=1 to k of (fi' * [product of all f_j with j<=k and not equal to i]***f_k+1**)} + f_k+1' (f_1 f_2 ... f_k). Lastly we notice that if we adjusted our sum to a sum from i=1 to k+1 the extra term we would get is exactly f_k+1' (f_1 f_2 ... f_k) and so we do this and finally get an equality to {sum from i=1 to k of (fi' * [product of all f_j with j<=k+1 and not equal to i]}.

That's it. A bit confusing not being able to write proper math notation and I went quickly so if you have any questions just ask.(11 votes)

- Would the answer be wrong if we used the product rule for the first two functions instead of the last two. In this case for f(x)g(x) instead of g(x)h(x)?(3 votes)
- No, you should get the same answer f´gh + fg´h + fgh´ either way.(4 votes)

- Two questions:

1) Is there some patterns for higher number of functions that can be summarized succintly?

2) Can this be used for derivative of binomials, like (x^+2)^3? Is there a pattern there?(1 vote)- The formula for differentiation of product consisting of n factors is

prod ( f(x_i) ) * sigma( f ' (x_i) / f(x_i) ) where i starts at one and the last term is n. Prod and Sigma are Greek letters, prod multiplies all the n number of functions from 1 to n together, while sigma sum everything up from 1 to n.

If you want to find the derivative of something in form let say (x^k + a)^n, then I would suggest for you just use the Chain rule, not Product rule.

Since you are going to be using chain rule very often in dealing with trigonometric, exponential and logarithmic expressions involving differentiation as well in other scenarios I would highly suggest you make sure you understand it.(3 votes)

- Does d/dx mean that the graph displays f(x) and does dy/dx mean that the y axis is the f(x)?

thanks for your help!(2 votes)- d/dx simply means that the derivative is taken with respect to x.(1 vote)

- Can someone explain in a simpler way? I have trouble with Algebra.(2 votes)
`F*(x)g(x)+F(x)g*(x)`

`First diff first function take second as it as + take first as it is and differentiate other`

(2 votes)

- How do we use this method to do derivation upto four terms? Please explain(1 vote)
- Let the four variables be a, b, c and d. Then we are trying to find (abcd)'. Break it up into 2 parts making use of the chain rule. Let x = a*b and y = c*d. Then we are finding (xy)' right? So,

(xy)' = x' * y + y' * x

= (ab)' * cd + (cd)' * ab

= (a'b+ab')cd + ab(c'd+cd')(2 votes)

- So, what would be the difference between using Sal's method here of the product rule, and leaving one of the terms the same and finding the derivatives of the other two?(1 vote)
- I am not completely sure what you mean by, " leaving one of the terms the same and finding the derivatives of the other two," but I think you mean rewriting it as f(x)[g(x)h(x)], and solving for just d/dx(g(x)h(x)). Then, using the product rule for f(x) times the result.

Well, What sal did*was*a little different from what you propose. Sal treated g(x)h(x) as one function*temporarily*but when he took the derivative, he only had to apply dy/dx to g(x)h(x), because of how the product rule works. If you were to take the derivative of just g(x)h(x) to start with, you are leaving f(x) out of the derivative. if you were to then take dy/dx ( f(x) ( g'(x)h(x) + g(x)h'(x) ) ), you would end up with second derivatives.

In other words you cannot take the derivative of part of an expression, and then use that to calculate the overall derivative. You must factor the entire expression into your calculations.

I hope this helps and that I explained clearly.(1 vote)

- why do you have to add it by its self again?(1 vote)
- Let's say you have a, b and c. You would first take the derivative of a and multiply that by b and c, then add all of that to the derivative of b multiplied by a and c, and lastly add the derivative of c multiplied by a and b. Visually it would look like this: (a')(b)(c) + (a)(b')(c) + (a)(b)(c'). You continue to add the products for each set for as many times as you have variables. So in the case of three variables, you would have three products to add together to get the final answer.(1 vote)

## 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.