[MUSIC] The correct rule for differentiating a product is f(x) = f(x) x g(x). We have that f prime (x) = f prime (x) x g(x) + f(x) x g prime (x). Therefore, we have that the derivative of a product of two functions is equal to the derivative of the first term times the second function plus the first times the derivative of the second. In Leibniz's notation, the product rule is expressed as d/dx[f(x) x g (x)] = [d/dx f(x)] x g(x) + f(x) x [d/dx g(x)]. If we use the product rule to find h prime of x where h of x is equal to (2x to the of power 6 + x top the power of 2) (7x to the power of 5- x to the power of 4). We have h(x) = f(x) x g(x) where f(x) = (2x to the power of 6 + x to the power of 2) and g(x) = (7x to the power of 5- x to the power of 4). The solution is f prime (x) = (12x to the power of 5 + 2x) and g prime (x) = (35x to the power of 4- 4x to the power of 3). Thus, h prime (x) = f prime (x) x g(x) + f(x) x g prime (x). Then we have (12x to the power of 5 + 2x) x (7x to the power of 5- x to the power of 4) + (2x to the power of 6 + x to the power of 2) x (35x to the power of 4- 4x to the power of 3). It is useful to simplify the answer by expanding in order to obtain just one polynomial. By multiplying each term, we get, h prime (x) = 154 x to the power of 10- 21x to the power of 9 + 49x to the power of 6- 4x to the power of 5. For example, we can use the product rule for differentiation in a simple economic framework. Let's D(P) to denote the demand function for product. By selling D(P) units, at price B per unit, revenues are R(P) given by R(P) = PD (P). From the economic theory, we know that normally, D prime (P) is negative. This is true because the demand for goods goes down when the price increases. According to the product rule for the differentiation, R prime (P) = D(P) + PD prime (P), which is the economic interpretation behind our derivative. If for instance B increases by one pound, the revenue R (P) will change for two reasons. Firstly, R (P) increases by 1 x D(P) because each of the T(P) will be worth one pound more. However, the one pound more in the price per unit will lead to change in the demand by D(P + 1)- D(P) units, which is about D prime (P). The loss due to a one pound more in the price per unit is therefore -PD prime P. This must be subtracted from D (P) in order to get R prime (P) as in our equation. The resulting expression we represent the fact that R prime (P), the total rate of change of R(P) is what the seller earn less what he lose. Let's differentiate a product of three factors. If we have y = (x + 4)(x + 6)(x + 3), y prime will be equal to d/dx(x + 4)(x + 6)(x + 3) + (x + 4) x d/dx (x + 6) (x + 3) + (x + 4)(x + 6) x d/dx (x + 3). And now we get, 1(1)(x+6)(x+3) + (x+4)(1) (x+3) + (x+4) (x+6)(1). Then we get x to the power of 2 + 3x + 6x +18 + x to the power of 2 + 4x + 3x + 12 + x to the power of 2 + 6x + 4x + 24. Which is equal to 3x to the power of 2 + 26x + 54. [MUSIC]