Proof of the product rule, the quotient rule, and the chain rule

In this post, I will prove the product rule, the quotient rule, and the chain rule, all of which are very important in finding derivatives.

Proof of the product rule

Here, I will prove the product rule,

First of all, consider a function, V(x), that is the product of two other functions, f(x), and g(x). So,

By the definition, the derivative of V(x) will be given by,

Now I will subtract and add f(x+h)g(x) in the numerator,

Now I will factorise in order to simplify the fraction,

Then, I will break up the limit like so,

I am going to break up the limit even further by writing it as,

The limit on the left will be equal to f(x), since h is tending towards 0. So, the equation will simplify to,

And, since g(x) is only a function of x, it can be taken out of it's limit,

Both the limits shown are, by definition, the derivatives of g(x), and f(x), respectively. Therefore,

And so, in conclusion,

The proof is now complete.

Proof of the quotient rule

Here, I will prove the quotient rule:

This can be easily proven by the product rule. First of all, let,

So,

And by the product rule,

Substituting g(x) back in, we get,

The only thing left, is to find,

This can be done using the chain rule, which will be proven next. According to the chain rule,

So, the equation will now simplify to,

Adding the fractions will give us,

Rearranging will give us,

Which is the quotient rule. The proof is now complete.

Proof of the chain rule

Here, I will prove the chain rule,

Let us work on the inner function, g(x), first. From the definition of the derivative, we know that,

I will use this as a basis of a function I am going to define, which will help prove the chain rule.

Let P(h) be a piece wise function defined in the following way,

Note that this function is continuous at h = 0, since the limit of the function approaches 0, as h approaches 0, and P(0) = 0.

Now, assuming h ≠ 0, we can rewrite the function in the following way,

I am going to do the same process for f(x), but I will use different variables so that the algebra later on does not get so confusing, instead of x, I will use y. The derivative of f(y) can be defined in the following way,

Now, let Q(k) be piece wise function defined in a similar way to P(h),

Q(k) is continuous at k = 0 in the exact same way that P(h) is, and by making the assumption that  k ≠ 0, I can rewrite this function in the following way,

Let's get on with the proof. By the definition, the derivative of f(g(x)) will be given by,

I am going to substitute the following (that I obtained earlier) in the equation,

So equation will now be,

Now, using this equation,

I am going to rewrite the limit again. I will let y = g(x), and k = h(P(h)+g'(x)), so therefore,

I avoided the substitution for Q(k) because that would look very messy and may be confusing to look at. Substituting this in my limit, I get,

The limit will simplify to get,

The h's will cancel out, so the limit will become,

Now, it should be obvious to see, that from the definition of P(h),

Also, remember that k, in this case, is equal to  h(P(h)+g'(x)), (I didn't write it in that way, because the equation would look too cluttered and confusing.) so,

And from the definition of Q(k),

So now the equation has simplified a lot. It will be,

Since g'(x), and f '(g(x)) are only functions of x, then,

And therefore,

The proof is now complete.