, 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.