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### Course: Differential Calculus>Unit 2

Lesson 10: Product rule

# Product rule

Discover the product rule, a fundamental technique for finding the derivative of a function expressed as a product of two functions. We'll learn how to apply this rule to simplify differentiation and enhance our understanding of calculus. Created by Sal Khan.

## Want to join the conversation?

• I was reading my calc book and they had (fg)' = fg'+gf' which is the reverse of what Sal was saying. I heard that it matters what order you take your derivatives. Does it?
(20 votes)
• In this case, no, because fg'+f'g=f'g+fg'. In that equation, because you're adding the two values together, you can move them around without worrying about getting a new value. If you were doing the quotient rule, though (another strategy when taking derivatives), the order would matter because of the subtraction sign between the two values: 2-3 does not equal 3-2, but 2+3 is equal to 3+2.
(120 votes)
• How do we assess whether to use the chain rule or the product rule?
(26 votes)
• The product rule is if the two "parts" of the function are being multiplied together, and the chain rule is if they are being composed. For instance, to find the derivative of f(x) = x² sin(x), you use the product rule, and to find the derivative of g(x) = sin(x²) you use the chain rule. See the difference?
(63 votes)
• Can Anyone Provide Me Proof For "Product Rule"??
(15 votes)
• Here's a short version. y = uv where u and v are differentiable functions of x. When x changes by an increment Δx, these functions have corresponding changes Δy, Δu, and Δv.

y + Δy = (u + Δu)(v + Δv) = uv + uΔv + vΔu + ΔuΔv

Subtract the equation y = uv to get

Δy = uΔv + vΔu + ΔuΔv

Divide through by Δx to get

Δy/Δx = u(Δv/Δx) + v(Δu/Δx) + Δu(Δv/Δx)

Now you can see where it's going, and the only problem is that last term at the end of the equation. We're given that v is differentiable, so as Δx approaches zero, the (Δv/Δx) part becomes the derivative of v, and because we're given that u is differentiable we know that u is continuous, and that means as Δx approaches zero, Δu approaches zero. So we have zero times a derivative that we know exists, which means we have zero, and we're left with the other terms, which give us the product rule when we apply the limit.
(30 votes)
• I am working with a Khan problem set. I see a simplification in one of the answers which is

1/4 log 4 = 1/2 log 2

I have reviewed the videos on log natural but I have failed to gain much intuition on the subject. How does LN work and why can it be simplified in this manner?
(5 votes)
• This is a log property, log (bᵃ) = a log (b)
¼ log 4
= ¼ log (2)²
= ¼ (2) log (2)
=½ log 2
This property works for any real log base, not just the natural log.
(20 votes)
• Can you do the chain rule in place of the product rule?
(4 votes)
• The two are not exactly interchangeable. There really is no way to evaluate the derivative of "x*sinx" with the chain rule. However, the two are often used in conjunction. If I had d/dx ( x*sin^2(x) ) I would use the product rule:
sin^2(x) * d/dx(x) + x * d/dx ( sin^2(x) ) = sin^2(x)*1 + x*2*sinx*cosx
Notice that I had to use the chain rule for d/dx( sin^2(x) ).
(13 votes)
• well I am confused
why can't we derivative directly like, what sal given is [x^2] sinx . why can't we write as
as 2xcosx.
please explain to me .
(4 votes)
• Whenever we have two functions being multiplied together, say f(x) and g(x), we have to use the product rule to differentiate their product

So, if we have h(x)=f(x)g(x), then to find h'(x) we would need to use the product rule which gives that h'(x)=f'(x)g(x)+f(x)g'(x)

To see why we can't simply differentiate h(x) as h'(x)=f'(x)g'(x), take the example where f(x)=x and g(x)=x, so h(x)=x*x=x^2

If we try to differentiate h(x) without the power rule, we'd get h'(x)=1*1=1, but that obviously isn't the case as we know that the derivative of h(x)=x^2 is h'(x)=2x.

Using the power rule on this example, we'd get h'(x)=x*1+1*x=2x, which agrees with what we got earlier
(7 votes)
• If d/dx of f(x)g(x) must have the product rule, isn't f(x)g(x)=(f*g)x and if so, can you apply the chain rule?
(5 votes)
• f(x) times g(x) is not the same thing as f of g of x. A simple example: Say that f(x) is x+1 and g(x) is x^2. While f(x)g(x) would be (x+1)x^2, f of g of x would be x^2+1. Continuing on with the same example, the f(x)g(x) derivative with the product rule would give x^2+2x(x+1), and the f of g of x derivative would be 2x. Clearly, not the same thing. Moral of the story: Just use the product rule when there are two functions being multiplied together.
(5 votes)
• How to prove the product rule?
(3 votes)
• what would be the derivative of cos^2x? Could anyone please explain how to get the answer? i'm a bit confused.
(3 votes)
• First, realize that cos² x is the same thing as (cos x)².
So, its derivative is:
2(cos x) ∙ d/dx (cos x)
We get this by applying the power rule and then the chain rule.
Now we apply d/dx (cos x) which is - sin x.
Thus, the derivative is:
2(cos x) (- sin x)
= - 2 (cos x)(sin x)
You can leave it that way or you use this identity:
sin (2x) = 2 sin(x) cos(x)
To simplify the derivative to: - sin(2x)
Thus d/dx (cos² x) = - sin(2x) = - 2(cos x)(sin x)
(4 votes)
• I'm still confused, for example what would dy/dx of y= (x2+1)^3(3x-5)^6

What would I use?
Can someone show me in steps?
(3 votes)
• Remember your product rule: derivative of the first factor times the second, plus derivative of the second factor times the first.

Your two factors are (x^2 + 1 )^3 and (3x - 5 )^6

So, you start with d/dx[ (x^2+1)^3 ] = 3(x^2+1)^2(2x) = 6x(x^2+1)^2 (Chain Rule!)
That gets multiplied by the second factor: 6x(x^2+1)^2(3x-5)^6

Now, do that same type of process for the derivative of the second multiplied by the first factor.

d/dx[ (3x-5)^6 ] = 6(3x-5)^5(3) = 18(3x-5)^5 (Remember that Chain Rule!)
That gets multiplied by the first factor: 18(3x-5)^5(x^2+1)^3.

Those two products get added together for your final answer:
6x(x^2+1)^2(3x-5)^6 + 18(3x-5)^5(x^2+1)^3

If you are taking AP Calculus, you will sometimes see that answer factored a little more as follows:
(x^2+1)^2(3x-5)^5[6x(3x-5) + 18(x^2+1)]
=(x^2+1)^2(3x-5)^5[18x^2-30x+18x^2+18]
= (x^2+1)^2(3x-5)^5(36x^2-30x+18)

Personally, I don't think I would normally do that last stuff, but it is good to recognize that sometimes you will do all of your calculus correctly, but the choices on multiple-choice questions might have some extra algebraic manipulation done to what you found.

Good luck! Hope it helped!
(3 votes)

## Video transcript

What we will talk about in this video is the product rule, which is one of the fundamental ways of evaluating derivatives. And we won't prove it in this video, but we will learn how to apply it. And all it tells us is that if we have a function that can be expressed as a product of two functions-- so let's say it can be expressed as f of x times g of x-- and we want to take the derivative of this function, that it's going to be equal to the derivative of one of these functions, f prime of x-- let's say the derivative of the first one times the second function plus the first function, not taking its derivative, times the derivative of the second function. So here we have two terms. In each term, we took the derivative of one of the functions and not the other, and we multiplied the derivative of the first function times the second function plus just the first function times the derivative of the second function. Now let's see if we can actually apply this to actually find the derivative of something. So let's say we are dealing with-- I don't know-- let's say we're dealing with x squared times cosine of x. Or let's say-- well, yeah, sure. Let's do x squared times sine of x. Could have done it either way. And we are curious about taking the derivative of this. We are curious about what its derivative is. Well, we might immediately recognize that this is the product of-- this can be expressed as a product of two functions. We could set f of x is equal to x squared, so that is f of x right over there. And we could set g of x to be equal to sine of x. And there we have it. We have our f of x times g of x. And we could think about what these individual derivatives are. The derivative of f of x is just going to be equal to 2x by the power rule, and the derivative of g of x is just the derivative of sine of x, and we covered this when we just talked about common derivatives. Derivative of sine of x is cosine of x. And so now we're ready to apply the product rule. This is going to be equal to f prime of x times g of x. So f prime of x-- the derivative of f is 2x times g of x, which is sine of x plus just our function f, which is x squared times the derivative of g, times cosine of x. And we're done. We just applied the product rule.