Product Rule The product rule. Examples using the Product and Chain rules.
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- Welcome back.
- I'm now going to introduce you to a new tool for
- solving derivatives.
- Really between this rule, which is the product rule, and the
- chain rule and just knowing a lot of function derivatives,
- you'll be ready to tackle almost any derivative problem.
- Let's start with the chain rule.
- Let's say that f of x is equal to h of x times g of x.
- This is the product rule.
- In the chain rule it was f of x is equal to h of g of x.
- I don't know if you remember that.
- In this case, f of x is equal to h of x times g of x.
- If that's the case, then f prime of x is equal to the
- derivative of the first function times a second
- function plus the first function times the derivative
- of the second function.
- Pretty straightforward.
- Let's apply it.
- Let's say that-- I don't like this brown color, let me pick
- something more pleasant.
- Maybe mauve.
- Let's say that f of x is equal to 5x to the fifth minus x to
- the seventh times 20x squared plus 3x to them mine 7.
- So one way we could have done it, we could just
- multiply this out.
- This wouldn't be too bad, and then take the derivative
- like any polynomial.
- But let's use this product rule that I've just shown you.
- So the product rules that says, let me take the derivative of
- the first expression, or h of x if we wanted to map
- it into this rule.
- The derivative of that is pretty straightforward.
- 5 times 5 is 25.
- 25x to the fourth, right?
- Then minus 7, x to the sixth.
- We're just going to multiply it times this second expression,
- doing nothing different to it.
- Maybe I'll just do it in a different color.
- Times 20x plus 3x minus 7.
- And then to that we will add the derivative of
- this second function.
- The derivative of that second function is 40x minus
- 21x to the minus eighth.
- And that times this first function.
- I guess I'll switch back to mauve, I think
- you get the point.
- 5x to the fifth minus x to the seventh.
- All we did here was we said OK, f of x is made of these two
- expressions and they are multiplied by each other.
- If I want to take the derivative of it, I take the
- derivative of the first one and multiply it by the second one.
- And then I add that to the derivative of the second
- one and multiply it by the first one.
- Let's do some more examples and I think that will
- hit the point home.
- Clear image.
- Change the colors and I'm back in business.
- Let me think of a good problem.
- Let me do another one like this, and then I'll actually
- introduce ones and the product rule and the chain rule.
- So let's say that f of x is equal to 10x to the third plus
- 5x squared minus 7 times 20x to the eighth minus 7.
- Then we say f prime of x, what's the derivative of
- this first expression.
- It's 30x squared plus 10x.
- And I just multiply it times this expression, right?
- 20x to the eighth minus 7.
- And I add that to the derivative of this second
- expression, this is all on one line but I ran out of space,
- 160x to the seventh, right?
- 8 times 20 is 160.
- And then the derivative of 7 is zero.
- So it's just 160x to the seventh times this
- first expression.
- 10x to the third plus 5x squared minus seven.
- There we go.
- And you could simplify it.
- You could multiply this out if you wanted or you could
- distribute this out if you wanted, maybe try to
- condense the terms.
- But that's really just algebra.
- So this is using the product rule.
- I'm going to do one more example where I'll show you,
- I'm going to use the product and the chain rule and
- I think this will optimally confuse you.
- I want to make sure I have some space.
- Here I'm going to use a slightly different notation.
- Instead of saying f of x and then what's f prime of x, I'm
- going to say y is equal to x squared plus 2x to the fifth
- times 3x to the minus three plus x squared to the minus 7.
- And I want to find the rate at which y changes relative to x.
- So I want to find dy over dx.
- This is just like, if this was f of x, it's just
- like saying f prime of x.
- This is just a Leibniz
- So what do I do in the chain rule?
- First I want the derivative of this term.
- Let me use colors to make it not too confusing.
- So what's the derivative of this term?
- We are going to use the chain rule first.
- So we take the derivative of the inside which is 2x plus 2
- and multiply times the derivative of the
- larger expression.
- But we keep x squared plus 3x there so it's times 5 times
- something to the fourth.
- And that something is x squared plus 2x.
- So there we took the derivative of this first term right here
- and then the product rules says we take the derivative of the
- first term, we just multiply it by the second term.
- So the second term is just 3x to the minus 3 plus x squared
- and all that to the minus 7.
- We did that and then to that we add plus the derivative of this
- second term times this first term.
- We're going to use the chain rule again.
- What's the derivative of the second term?
- I'll switch back to the light blue.
- Light blue means the derivative of one of the terms.
- So we take the derivative of the inside, the derivative of
- inside is minus 3 times 3 is minus 9, x go down one to
- the minus 4, plus 2x.
- And now we take the derivative of the whole thing.
- Times minus 7 times something to the minus 8, and that
- something is this inside.
- 3x to the minus 3 plus x squared.
- And then we multiply this thing, this whole thing which
- is the derivative of the second term times the first term.
- Times, and I'm just going to keep going, times x squared
- plus 2x to the fifth.
- So this is a really, I mean you might want to
- simplify at this point.
- You can take this minus 7 and multiply it
- out and all of that.
- But I think this gives you the idea.
- And if you have to multiply this out and then do the
- derivative if it's just a polynomial, this would
- take you forever.
- But using the chain rule, you're actually able to, even
- though we ended up with a pretty complicated answer,
- we got the right answer.
- And now we could actually evaluate the slope of this very
- complicated function at any point just by substituting the
- point into this fairly complicated expression.
- But at least we could do it.
- I think you're going to find that the chain and the product
- rules become even more useful once we start doing derivatives
- of expressions other than polynomials.
- I'm going to teach you about trigonometric functions and
- natural log and logarithm and exponential functions.
- Actually, I'll probably do that in the next presentation.
- So I will see you soon.
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