In today's video, we apply the Product Rule to further explore the behavior of the Gaussian curves y equals e to the minus x squared using the second derivative, and to give the proof by mathematical induction, of the formula for differentiating a power of x with the exponent as a positive integer. In an earlier video, we use methods of curve sketching to go a long way towards describing the graph of the function y equals e to the minus x squared. We observed, that the y-intercept is 1, that there are no x-intercepts, and the x-axis is a horizontal asymptote. We used the Chain Rule to find the derivative y dash and built its associated sign diagram. We put this all together, noting the y-intercept, the asymptotic behavior of the x-axis both to the far left, and far right and in fact that there is a turning point when x equals 0. Because, the curve is increasing to the left and decreasing to the right, it's natural to fill in the rest of the curve to obtain this bell-shape. To created this shape there appear to be true natural inflections where the curve changes concavity. To be sure about the way the concavity changes, we need to go further and investigate the second derivative y double dash. Because, the first derivative is quite complicated, we'll use the product rule to work out its derivative. We have y dash is minus two x times e to the minus x squared, which we can write as u times v, where u is minus 2x and v is e to the minus x squared. Then u dashed is minus 2, and v is our original y, so v dash is in fact the same as y dash. By the product rule, y double dash which is the derivative of y dash, is uv dash plus vu dash. Which is this expression, which becomes after a couple of steps 2 times 2x squared minus 1 times e to the minus x squared. Notice, that the factor e to the minus x squared is always positive. Hence, the second derivative is zero precisely when 2x squared minus 1 equals 0. That is, when x equals plus or minus 1 over the square root of two, and we can build the sign diagram, noting the important points for x where y double dash is 0, and the pattern of plus minus plus, producing a pattern of concave up, concave down, concave up, as we pass from left to right. This tells us that there are inflections when x equals plus or minus 1 over root 2. We can update our previous information in sketching the Gaussian curve, confirming what we expected the concavity might be, and to locate the points of inflection. This now completes a thorough investigation of the curve. Our second application of product rule, is to provide a prove of the formula for the derivative of x to the n. You will recall that we carefully proved, from the limit definition that the derivative of x squared is 2x, and the derivative of x cubed is 3x squared. This is a special case of a general pattern that the derivative of x to the n is, n times x to the m minus 1. A result that were used on several occasions without explaining why it's true. The aim of this next application is to carefully prove this formula for all positive integers n, that is for n equals 1, 2, 3, 4, and so on. The argument that I'm about to give is called more formally a proof by mathematical induction. Which is a very common and powerful proof technique in mathematics which you can read about if you wish, though I think you'll be able to follow the main ideas. For many of you, this might be your very first proof by mathematical induction, and will serve you well as a prototype example of the technique, in case you go on to do more higher mathematics. The claim that we are proving is in fact infinitely many statements as n passes through the set of positive integers. We've already carefully checked this in the cases n equals 1, 2, and 3. The case n equals 1 matches up nicely to the fact that the derivative of x is 1. The case as n equals 2 and 3 match up with facts, we already know. The problem for us now is to somehow prove the cases n equals 4, 5, 6, and so on. We don't want to spend a lot of time and effort on each case, and we certainly don't have an infinite amount of time to go through each and every positive integer separately. This is where you might start, to see the power of thinking abstractly. What follows is like a thought experiment. Suppose, that by whatever means, we've been able to prove the formula for some particular positive integer n, and let's refer to this formula, only for this particular n, by star. In formal proofs the statement star is referred to as the inductive hypothesis. Our aim is to prove statements star with n replaced by the next positive integer n plus 1. This new statement would be that the derivative of x to the n plus 1 is n plus 1 times x to the n, we can refer to this as double star. If we can prove that star implies double star, then we get an instantaneous infinite chain reaction from the case n equals 1, and then star must hold for all positive integers. Why are we able to say that? Well, we've verified star for n equals 1, and also for n equals 2 and 3, but that's not in fact important. It's enough that we're verified it for n equals 1. Star implying double star means the formula must be true for 1 plus 1 equals 2, and then feeding two back into n for 2 plus 1 equals 3, and then feeding three back into n for 3 plus 1 equals 4, and repeating for 4 plus 1 equals 5, for 5 plus 1 equals 6, and so on forever racing through all the positive integers instantaneously. So, let's prove double star after supposing that star is true for some particular positive integer n. Notice that x to the n plus 1 is the product of x to the n with x. So, we can apply the product rule to conclude to the derivative is x to the n times the derivative of the second factor, plus x times the derivative of the first factor. Which becomes x to the n times 1, plus x times, and at this point we invoke statement star n times x to the n minus 1. Then this tidies up in a few steps to become n plus 1 times x to the n. This establishes the statement double star hence, star implies double star. This implication is called the inductive step, in formal proofs by mathematical induction. It shows by the chain reaction effect that I mentioned earlier, that star holds for all positive integers n. This completes this particular application of the product rule for all positive integers n and for all real numbers x. If we restrict attention to positive real numbers x only, then we can go much further and prove that to differentiate a general power, say x to the Alpha for any real number Alpha, you again bring the exponent to the front and make a new exponent by subtracting one. This very general fact follows from the chain rule and properties of logs and exponentials. To see why, express x to the Alpha as e to the Alpha times ln of x, and then the derivative of x to the Alpha becomes, the derivative of e to the u, where u equals Alpha times ln of x. Which becomes d/du of e to the u times du/dx by the chain rule and each piece is then straight-forward. The derivative of e to the u with respect to u, is just e to the u and the derivative of u with respect to x, is just Alpha over x because the derivative of ln of x is 1 over x. Thus the derivative becomes e to the u times Alpha over x, which can be rewritten as x to the Alpha times Alpha times x to minus 1. Which simplifies Alpha times x to the Alpha minus 1, and this completes the proof of our formula. In today's video, we applied the product rule first to complete our curve sketching analysis of the Gaussian curve from an earlier video, in particular to understand the curve's behavior with respect to concavity and to locate the inflection points. Secondly, to prove that the derivative of x to the n is n times x to the n minus 1 and for all real numbers x, and all positive integers n. This proof in fact is an example of the technique of proof by mathematical induction. We then used the chain rule and properties of logs and exponentials to see how to generalize this formula for arbitrary real exponents. 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.
