In today's video, we introduce, explain and illustrate the Product Rule, which enables you to differentiate a product of expressions in terms of the derivatives of the factors. Consider a function y of x which is a product of u and v. Where u and v are themselves functions of x. The product rule states that the derivative of y with respect to x is u times the derivative of a plus v times the derivative of u, expressed this way using Leibniz notation. Equivalently, using function notation and the dash to denote the derivative, the product rule says y dashed equals uv dashed plus vu dashed. For example, let y be this product of a quadratic and a linear polynomial and we want to find the derivative of y dash. We'll give two solutions. The first solution is direct without using the product rule. We expand the brackets and gather together the terms to form this cubic polynomial and they differentiate immediately to get the quadratic 6x squared plus 2x minus 23. The second solution uses the product rule. We write y as the product uv where u equals the first factor, x squared plus 3x minus 4 and v equals the second factor 2x minus 5, so that the derivative of u is 2x plus 3 and the derivative of v is 2. But the product rule, y dash equals uv dash plus vu dash and we just put all the pieces together. Expand and simplify and we get the quadratic 6x squared plus 2x minus 23, which of course agrees with our first solution. The solutions are the same and the pathway is quite different. Which pathway you take depends on the problem at hand. Here is another example. Again, y is a product of two expressions involving x and we want to find the derivative y dash. We can solve this directly by expanding the brackets. Simplifying one of the expressions using an exponential law and then differentiating directly piece by piece, using the fact repeatedly that the derivative of e to the kx is ke to the kx where k is a constant, which is a result we established in an earlier video using the chain rule. Then tidying this up a little to get negative 3e to the 3x minus 2e to the negative x minus 2e to the 2x. An alternative solution is to apply the product rule recognizing that y is the product uv, where u the first factor, one plus e is to the minus x and v is a second factor, two minus e to the 3x. So, that u dash quickly becomes negative e to the negative x and v dash becomes negative 3e to the 3x. By the product rule, y dashed equals uv dash plus vu dash. We can put all the pieces together, expand, simplify and gather terms to get finally negative 3e to the 3x minus 2e to the 2x minus 2e to the negative x, which agrees with our first solution. Using the product rule is often the only natural way to proceed, and of course, one tries to avoid going back to the original limit definition of the derivative. Here's such an example. We want to find the derivative y dash, when y is the product of x sine x. We'll present the solution using the product rule and Leibniz's notation. The derivative is dy-dx which is d-dx of x sine x, which by the product rule, is the first factor x times the derivative of the second factor plus the second factor, sine x times the derivative of the first factor. The derivative of sine x is cos x and the derivative of x is one, so we get x cos x plus sine x times one, which is simply x cos x plus sine x. Notice how just in a few short steps, we are able to painlessly differentiate a sophisticated function. You're very unlikely to guess this answer and extremely unlikely to navigate through the limit definition of the derivative, which would quickly become a quagmire in an example like this. You can appreciate the power of results like the product rule that provides such succinct and elegant solutions. We've only stated the product rule and used it and you might be curious to know why it works. Here's a sketch of a proof why it works and what follows now is quite advanced, tricky and intricate. So, you shouldn't worry if you find it difficult to follow all of the details. Let y be a product of u and v where y, u and v are functions of x. A small change in x called Delta_x propagates a small change in y called Delta_y. Well, Delta_y is the difference between y evaluated at x and y evaluated at x plus Delta_x. That is, Delta_y equals y of x plus Delta_x minus y of x. But y is u times v, so this becomes u of x plus Delta_x times v of x plus Delta_x, minus u of x times v of x. Now comes the magic sleight of hand. We insert minus u of x plus Delta_x times v of x, plus u of x plus Delta_x times v of x, which altogether is zero, so it doesn't change the overall value of the expression. Now, why would we do such a thing making it look so much more complicated? Well, it's common in proofs to mathematics to introduce terms to make things look more complex or elaborate, but to facilitate something that gets you past an obstruction or leads to something ultimately more simple than what you started with. The trick here, using an expansion of zero expressed as an addition and subtraction of the same thing, has lots of applications in algebra. You might remember we employed a similar trick to complete the square when manipulating quadratics in earlier videos. Observe that in the first two terms of this expression is a common factor of u x plus Delta_x, which we can bring outside of v of x plus Delta_x minus v of x. In the second half, there is a common factor of v of x which we can bring outside on the right of u of x plus Delta_x minus u of x. Which is significant progress because v of x plus Delta_x minus v of x is just Delta_v, the change in v and u of x plus Delta_x minus u of x is just Delta_u, the change in u. Now, you start to see how everything is starting to simplify. We have Delta_y equals this combination and we can divide everything through by Delta_x, setting up something that looks close to the derivative dy-dx. The left-hand side, Delta_y over Delta_x approaches the derivative dy-dx as Delta_x approaches zero. We can see what happens to each piece on the right-hand side. We have u of x plus Delta_x approaches u of x as Delta_x approaches zero, in fact, using a property called continuity and throughout this discussion, all functions turn out to be continuous as a consequence of assuming that the derivatives exist which is a subtle point explained in more advanced courses in calculus. Delta_v over Delta_x approaches dv-dx, Delta_u over Delta_x approaches du-dx and v of x just remains as v of x. We can rewrite the right-hand side as u of x du-dx plus v of x du-dx. Or in function notation y dashed equals uv dashed plus vu dashed and we've finished the proof of the product rule. In today's video, we introduced and illustrated the product rule which enables you to differentiate a product of expressions in terms of the derivatives of the factors. We then sketched a proof which involved some subtle algebraic manipulation and the limit definition of the derivative using Leibniz notation. Please read the notes and when you're ready, please attempt the exercises. Thank you very much for watching and I look forward to seeing you again soon.