Calculus - Weekly Reflection #4
So this week in Calculus was... fairly simple and to the point. Things we learned include taking the first derivative, using the first derivative test, taking the second derivative, using the second derivative test, finding intervals on which functions increase and decrease, finding max's and min's of a function, and justifying our answers in Calculusspeak. :-) (yes, Calculusspeak).
Anyway, this coming Monday we have a test on everything we've learned so far in Calculus. EVERYTHING. So..., I think I'm going to help some of you out with a problem that might come up. Some people are getting confused when using the Chain Rule with trigonometric functions. The Chain Rule, if you don't know it, is when you have something like (2x+4)3. For this, you would bring the 3 to the front, copy the 2x+4 and make it squared, then times by the derivative of the inside (that's the chain rule part) which is 2 to make it a whopping end function looking like 6(2x+4)2. Right? That's all fine and dandy. Some people may understand that part just fine, which is good, but for some reason they don't understand it when you throw in a trigonometric function. So let's look at this:
sin3(x2). Some people are thinking to themselves, uh-oh. sin3? We don't have a derivative formula for that. Well you are wrong. :-) The way to help with this type of problem is simply this. Rewrite the previous as this:
(sin(x2))3 Now--doesn't this look REALLY similar to the problem we did before? Bring the 3 to the front...make it squared...and times by the derivative of the inside (yes, i love me some chain rule!) and we wind up with:
6x(sin2(x2)) cos(x2)
Oh, are you saying "well john, we brought a 3 to the front, how in the heck did you get a 6x?" Well, duh! Chain rule on the derivative of cos(x2). Have to times by the derivative of the inside (which is 2x).
Well I hope that helps out you guys that may be getting confused and lost on those type of derivatives.
Now, for something I need help with...
Can someone explain to me the steps and purpose of the First Derivative Test? I mean, I was sick the day we learned it, and I've tried to grasp the concepts of it but I'm not exactly sure on what I'm doing with it or why for that matter. I think maybe it can be used to find maxs and mins, but I'm unsure! Any help would be great guys. :-)
Study study study for the quiz tomorrow. :-)
-John
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well john i'm working on our homework now and i believe i am doing this right. So as far as i know, when you have an original function, you take the first derivative because you are going to find your critical points. You take the derivative and set it equal to zero. Then you solve for x. Whatever number or numbers you get for x that is your critical points. You use the critical points to set up intervals to see where the function is increasing or decreasing. Then if you want to find the max and min, you plug a number in between the two intervals into the derivative function.
ReplyDeletefor example: if an interval is (-infinity, 2) you would choose 1 to plug into the derivative equation, when you get that number,it will tell you whether that number is a max or a min. Hope this helps!
good explanation but don't forget to include where a function is not continuous...
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