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