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, How do we prove the quotient rule? Well
first, we should remember what the

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quotient rule says. So, remember what the
quotient rule says. It says the derivative

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of f over g is the derivative of f times g
minus f times the derivative of g all over

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the value of g. Not the derivative of g.
Just g of x squared. But we haven't

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actually seen a proof of the quotient
rule. Why should the derivative of a

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quotient be governed by that crazy looking
formula? Well, one way to justify this

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formula is to combine the chain rule and
the product rule. So, we're trying to

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build our way up to the quotient rule so
we can first do the simplest possible case

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the quotient rule by hand if you like.
What's the derivative 1 over x? Well, 1

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over x is just x to the minus first power.
And if I differentiate this, that's the

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power rule. We saw how to do that before.
That's minus 1 times x to the minus second

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power. Another way to write this is minus
1 over x squared. Now, if you weren't

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certain why the power rule held, you could
also have calculated this derivative by

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hand by going back to the definition of
derivative. This is actually how we

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justified the power rule for negative
exponents. This limit, you can also

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calculate, is minus 1 over x squared.
Knowing how to differentiate 1 over x is

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enough for us to differentiate 1 over g of
x by using the chain rule. So, we're going

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to use the chain rule. So, let me first
make up a new function that's called f of

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x at function 1 over x. So then f prime of
x is minus 1 over x squared. The

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derivative that we just calculated. Now,
if I wanted to calculate the derivative of

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1 over g of x. Well, that's the same as
the derivative now of f of g of x, since I

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defined f to be the 1 over function. And
by the chain rule, the derivative of this

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composition is the derivative of the
outside at the inside times the derivative

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of the inside. The derivative of f is
minus 1 over its input squared. So, f

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prime of g of x is minus 1 over g of x
squared. That's looking good. That's like

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th e denominator of the quotient rule.
Alright, times g prime of x. so, the

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derivative of 1 over g of x is minus 1
over g of x squared times the derivative

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of g. Now, I've got the product rule. So,
if I can differentiate f and I can

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differentiate 1 over g of x, I can
differentiate their product which happens

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to be the quotient, f of x over g of x.
So, I want to differentiate f of x over g

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of x, right? I'm trying to head towards
the quotient rule. But I'm going to

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rewrite this quotient as a product. It's
the derivative of f of x times 1 over g of

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x. Now, this is the derivative of products
so I can apply the product rule, which I

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already know. And that's the derivative of
the first times the second, plus the first

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times the derivative of the second. But we
calculated the derivative of the second

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just a moment ago. The derivative of 1
over g of x is minus 1 over g of x squared

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times g prime of x. So, I can put that in
here as the derivative of 1 over g of x.

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It's minus 1 over g of x squared times g
prime of x. By rearranging this, I can

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make this look like the quotient rule that
we're used to. Let's aim to put this over

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a common denominator. So, I could write
this as f prime of x times g of x over g

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of x squared plus, what do I have over
here? Well, negative f of x g prime of x,

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the negative 1 f of x g prime of x, over g
of x squared. And I can combine these two

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fractions, f prime of x g of x minus f of
x g prime of x all over g of x squared.

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That's the quotient rule, right? We've got
to the quotient rule at this point. And

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how is it we do it? Well, think back to
what just happened. I, I used the power

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rule to differentiate 1 over x. Once I
knew the derivative of that, I could use

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the chain rule to differentiate 1 over g
of x. And once I knew the derivative of 1

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over g of x, I could then use the product
rule on f of x times 1 over g of x to

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recover the quotient rule. What's the
upshot here? Why is it important that the

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quotient rule can be seen as an
application of the chain rule and the

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product rule? One reason is a pedagogical
one. I think it's important for you to see

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that all these differentiation rules are
connected together. I hope that will give

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you a better sense of the rules and, and
make them more memorable.