**EDIT: So, apparently, I forgot to post this yesterday. Whoops.
What a weird week. From #Peppergate to winning at Quizbusters, this has been an eventful, if not slightly confusing, week. And what we are learning in math has definitely contributed to the confusion. The topics of this week's learning: "u" substitution and implicit differentiation! Because of the whole pepper spray incident, we didn't get to do a whole lot on Monday, but we started off Tuesday learning how to anti-derive chain rule functions using "u" substitution, which involves identifying a part of the derivative as "u" and using substitution to come out with the anti-derivative. If, for example, S 6x^3(3x^4-8)^3 dx then, to solve for the anti-derivative: 1. Define "u" in terms of x ("u" should be the "inside") u=3x^4-8 2. Find the derivative of "u" in terms of x du=12x^3 dx 3. Solve so that du can be substituted into the derivative du/12= x^3 dx 4. Substitute "u" and "du" into the derivative S 6(3x^4-8)^3*x^3 dx S 6u^3 du/12 5. Anti-derive that thing you got in the last step. 1/12*6u^3 du 1/8u^4+c 6. Substitute "x" back in. 1/8(3x^4-8)^4+c Tada!! (Hopefully I did that right; I made the example up). That concept took up the first half of our week. It wasn't particularly difficult to learn, but sometime I lost track of what parts I was deriving and what parts I was anti-deriving, which was a bit of a nuisance. But, like I said earlier, not too hard. The next thing we did was implicit differentiation, or finding the derivative when y isn't explicitly defined, like when y^2=x. I'm not gonna go into too much detail explaining this process (the picture has my notes on this stuff, if you really want to see it), but basically, differentiate y with respect to x by using the chain rule, factor out dy/dx, and then solve for dy/dx. That's the gist of it. Again, like most of the stuff we learned these past few weeks, this concept wasn't all too difficult to grasp. My only problem is that there are times when I forget when and what I'm supposed to be differentiating. You can get a loooonng stream of numbers in implicit differentiation. I'm talking about longer than the quotient rule long. It can get kinda confusing keeping the numbers straight, so I have to go back and check my work to make sure I didn't make any mistakes. Which, you know, I should probably do anyway. Procrastination.
I should stop doing that. It's really not good for my mental health. Anyways, this week we discussed, surprise, surprise, derivatives! Apparently there even more properties of derivatives to be learned! (That was not sarcasm, I promise you.) Our focus this week was on the derivatives of composite functions and the chain rule. As we discovered through part of our activity on Monday, the derivative of a composite function at a point equals the composition of the equations of the tangent lines. The chain rule is used to the derivative of such functions; it states: the derivative of a composite function f(g(x)) equals f'(g(x))*g'(x). Essentially, the derivative is the derivative of the outside function with the inside unchanged multiplied by the derivative of the inside function. This concept isn't particularly hard to understand, but I had a somewhat difficult time learning it because I was absent the day of this lecture and activity. That tends to be a hindrance on one's learning. However, thanks to Cresswell's completely legible and not at all incoherent notes, I was able to catch up quickly. In all seriousness, though, I had only a few problems understanding the chain rule but as soon as I got the notes, I got it. Also, there are instances where two or more chain rules have to be done, like in the case of f(x)=(cos(3x^2))^2, which also made life a bit harder. The first step to solving this problem, or at least, the first step that helps me, is to identify the different functions that makes up the composition. For f(x), it would be x^2 (the outermost layer), cos(x) (the middle layer), and 3x^2 (the innermost layer). Once all parts have been identified, just follow the chain rule! Take the derivative of the outermost layer, leaving the inside intact: f'(x)=2(cos(3x^2)) Then multiply that by the derivative of the next innermost layer: f'(x)=2(cos(3x^2))*(-sin(3x^2)) Finally, multiply that by the derivative of the ultimate innermost layer: f'(x)=2(cos(3x^2))*(-sin(3x^2))*(6x) And that's your derivative! It seems like a long and complicated process, but once the basics are down, it is relatively easy. I just make it hard on myself by *sigh* procrastinating. Two posts in one week messes up my naming system.
This post is about derivatives and epiphanies about derivatives. Last Monday, 10th of October, we did an activity in class that had us explore transformations of functions and their effects on the derivatives of those functions. It turns out that reflections and vertical stretches affect both the function and its derivative. For example, say f(x)=x^(1/2) with a derivative of f'(x)=1/2*x^(-1/2). If f(x) is vertically stretched by a factor of 2 and it is reflected over the x-axis, the new function would be f(x)=-2x^(1/2). What we learned in this activity is that the transformations done to the function will be the same transformations done to the derivative. The derivative of the new function is f'(x)=-x^(-1/2), which is the derivative of the original function vertically stretched by a factor of 2 and reflected over the x-axis. Although in the activity we only did this with square root functions, this property applies to all types of functions because square root functions can be treated like any other exponential function. For example, if g(x)=2x^3 and it is vertically stretched by a factor of 3 to become g(x)=6x^3, then its original derivative, g'(x)=6x^2 will become g'(x)=18x^2. The new derivative is the old derivative vertically stretched by a factor of 3. I tried to make a GIF of a graph demonstrating this concept, but that failed miserably; I have instead linked it. What is the derivative of a function? It's the slope of the tangent line at any point, of course! Well, as you have probably surmised, this week's focus was derivatives. Hooray! Apparently we had learned a bit about derivatives before, especially concerning how to calculate the derivative of a function at any point (more on that later), in Honors Pre-Calc. However, instead of calling it derivatives like any normal thing would do to save confusion, the book called it the difference quotient. (Why do you do this to us, math textbooks? *shakes fists at sky*) Our first activity with derivatives this week dealt with what I like to call the "zooming in" characteristic of derivatives. The official term is local linearity. (I like mine better, though) In this activity, we discovered that by zooming closely into a curve a point, the curve would appear to be a straight line in the graphing window. We could then calculate the slope of the line in respect to a certain point. Here we were also shown a few situations where the derivatives do not exist, a topic which was elaborated on later in the week. The derivative fails to exist if when x=a, there is a corner, cusp, vertical tangent, or discontinuity. This activity led us to the formal definition of the derivative of a function with respect to x: Which we all learned in Pre-Calc as the difference quotient. See rant in first paragraph. It uses the idea of limits that we discussed last week and simplifying skills learned in Alegebra II to create a function that describes the slope of the tangent line at any point on the graph. Anyways, we used the formula to solve a few problems, but it wasn't that hard. There was just a ton of simplifying. But I have no complaints. I like to simplify expressions. No sarcasm at all. Our next topic, though, focused less on finding the derivative more on the properties of the derivative function's graph in comparison to the original function. It turns out that where the original function had a rising slope, the derivative will be above the x-axis. If it was a falling slope, the derivative would be under the x-axis. The maximums and minimums of the original function turn into the zeroes of the derivative. What fascinates me is that, although we can predict the graph of a derivative based on these properties, the inverse cannot happen. The graph of the original function cannot be obtained from the graph of its derivative because, if given only the derivative, the information is not specific enough.
The rest of the week focused on solidifying these ideas into our heads. (Or beating it into our skulls, whatever way you prefer.) I didn't find this subject particularly hard, mainly because we covered this a bit in Pre-Calc. The concepts were also relatively simple to get as well. I also found some really cool (albeit a bit cheesy) videos that NOVA created relating derivatives to surviving the zombie apocalypse. These are the videos that I've been meaning to share with Mr. Cresswell, but I keep forgetting to email him. Welp, they're here now. Now that we are getting out of the review stage of Pre-Calc and into actual Calculus, I'm getting more and more excited for this class!! |