This week in calculus, we went over a lot. We had to remember finite limits, infinite limits, and points of discontinuity from last year. We also went over some new material. The new things we went over were average speed (or slope), instantaneous speed, and derivatives. We were given derivative formulas to help us and give us shortcuts for the lengthy formulas we learned for average speed and instantaneous speed.
I learned that when working average speed, you will be given a word problem, and in that word problem there will be either two numbers or a hint (in the first two seconds) to two numbers. You will also have an equation at the end of the word problem (y=16t^2). These numbers in the word problem will become the x variables in your formula. To get the y variables, you plug both the x variables into the equation to get two separate ys. Once that is done, you plug your xs and ys into the midpoint formula (because midpoint finds slope/average speed). The midpoint formula is y2-y1/x2-x1. Once you get your number, you are to write the units next to it (unit length/unit time).
I also understand the product rule formula in our derivative formulas. If you are asked to take the derivative of a product you are to copy down the first term being multiplied times the second term's derivative plus copy down the second term and times that by the first term's derivative.
For the most part, I believe I understand all the concepts of derivatives so far. I just get mixed up when a formula has a fraction, such as the quotient rule, or if a problem ends up having a negative exponent. When I look at the problem, I always want to put the variable with the negative exponent under everything, instead of just what it's supposed to be over. I also confuse myself when I see a number minus a number times something else. Such as 5-3sinx. When I see this, I automatically think I can combine these two numbers into one.
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Good job Mamie
ReplyDeleteOkay, as far as I know, you leave the exponent negative, unless it is like this example below.
ReplyDelete1) Original Problem: 5/x^2
2) Bring the 5 down and make the exponent negative: 5x^(-2)
3) Multiply the constant by the exponent and minus 1 from the exponent: -10x^(-3)
4) Take the 10 and put it back on top (where the 5 was) and take the negative off of the exponent.
As for your 5-3sinx problem, I'm pretty sure you just (if this is what you're doing in that problem) take the derivative of 5 (0) and -3sinx (-3cos * 1 = -3cosx) and your answer would be -3cosx.
I hope that's right and I helped. :)