[MUSIC] From your calculus course, you know how to compute the derivative of the composition of two functions. This is done by the chain rule. Essentially the same rule holds for a formal power series. Suppose we have two power series A(q). A0 + a1q + a2q squared plus etc. And B(q), and B(q) has no constant term so it starts with b1q + b2q squared + b3q to the 3rd plus etc. And if you want to compute the derivative of their composition. So theorem chain rule says that A(B(q)) prime is A prime of B(q) times B prime of q. Proof. And straightforward, so let's take A(B(q)) prime and this is the derivative of anB(q) to the power n. And the whole thing is linear. So the derivative operator is linear. So, this is equal to anB(q) to the power n prime. And this thing can be computed by the latence rule. So, this is, And greater than or equal to 0, and when you apply the latence rule to this product, you get, an, and then you get n times B prime of q, times B(q) to the power of n- 1. Because you apply the derivative of the product of n factors and each of the factors gives you the derivative of each factor is B prime and the rest is B to the power n- 2. So in each column you have, B prime of q. And here you have an times n times B(q) to the power n- 1. And B prime of q. And this is nothing but the derivative of, The power series A composed with B. So this is A prime (B(q)) times B prime of q. The proof is finished. A very important corollary of this chain rule is the inverse function theorem. Which is also probably familiar to you from the MLS' course. So if we have two powers of series which are inverse to each other, so if A of B(q), if their composition is just q then the derivatives are related as follows. Then A prime (t) is equal to 1 over B prime of q, Where t is B(q). The proof is as follows. Let us take this equality and compute the derivatives of both sides using the chain rule. That means that A(B(q)) prime times B prime of q is q prime, that is 1. So, A prime of B(q) is 1 over B prime of q. So A(t) is 1 over B prime(q) where t is B(q). The inverse functional theorem is proved. We'll use the theorem to define logarithm. [MUSIC]