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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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