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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Â